Unit - 3

Higher order linear differential equations with constant and variable coefficients

Higher order linear differential equations with constant coefficients

Second order linear homogeneous equations with constant coefficients

A homogeneous linear ordinary differential equation with constant coefficients is an ordinary differential equation in which coefficients are constants, all terms are linear, and the entire differential equation is equal to zero,

The form of second order linear differential equation with constant coefficients is,

Where a,b,c are the constants.

Let, aD²y+bDy+cy = f(x), where d² = d2/dx2 , D = d/dx

∅(D)y = f(x) , where ∅(D)y = aD²y+bDy+cy

Here first we solve, ∅(D)y = 0, which is called complementary function(C.F)

Then we find particular integral (P.I)

P.I. = f(x)

General solution = C.F. +P.I.

Let’s do an example to understand the concept,

Example1: Solve (4D² +4D -3)y = e2x

Solution: Auxiliary equation is 4m² +4m – 3 = 0

We get, (2m+3) (2m – 1) = 0

m = -3/2, 1/2

Complementary function: CF is A e-3x/2+ B e1 x/2

Now we will find particular integral,

P.I. = f(x)

General solution is y = CF + PI

= A e-3x/2+ B e1 x/2 + 1/21 . e2x

Non-Homogeneous equations of higher order with constant coefficient of R.H.S. Terms of the type eax, sin ax, cos ax

Linear differential equation is those in which the independent variable and its derivative occur only in the first degree and are not multiplied together.

Thus, the general linear differential equation of the n’th order is of the form

Where p1, p2, . . ., pn and X are function of x.

Linear differential equation with constant co-efficient are of the form-

Where k1, k2, . . . , kn are constants.

Rules to find the complementary function-

To solve the equation-

This can be written as in symbolic form-

(dn + k1dn – 1 + k2dn – 2 + . . . + kn)y = 0 ………….(2)

Or-

(dn + k1dn – 1 + k2dn – 2 + . . . + kn) = 0

It is called the auxiliary equation.

Let m1, m2, . . . , mn be the roots-

Case-1: If all the roots are real and distinct, then equation (2) becomes,

(D – m1)(D – m2) . . . (D – mn)y = 0

Now this equation will be satisfied by the solution of (D – mn)y = 0

This is a Leibnitz’s linear and I.F. =

Its solution is

The complete solution will be-

Case-2: If two roots are equal m1 = m2

Then complete solution is given by-

y = (c1x2 + c2x + c3)

Case-3: If one pair of roots be imaginary, i.e. m1 = α + iβ, m2 = α - iβ then the complete solution is-

y = eax (c1 cos β x + c2 sin β x) + c3

Where C1 = c1 + c2 and C2 = i(c1 – c2)

Case-4: If two points of imaginary roots be equal-

m1 = m2 = α + iβ, m3= m4 = α - iβ

Then the complete solution is-

y = eax[(c1 x + c2) cos β x + (c3x + c4) sin β x)] + . . . + cn

Example-Solve (D2 + 6D + 9) = 0

Sol.

Its auxiliary equation is-

D2 + 6D + 9 = 0

(D + 3)2 = 0

Where-

D = -3, -3

Therefore, the complete solution is-

x = (c1 + c2t)e-3t

Inverse operator-

is that function of x, not containing arbitrary constants which when operated upon gives X.

So that satisfies the equation f(D)y = X and is, therefore, its P.I.

f(D) and 1/f(D) are inverse operator.

Note-

1.

2.

Rules for finding the particular integral-

Let us consider the equation-

Or in symbolic form-

(dn + k1dn – 1 + k2 dn – 2 + . . . + kn)y = X

So that-

P.I. = 1/(dn + k1dn – 1 + k2 dn – 2 + . . . + kn) X

Now-

Case-1: When X = eax

In case f(a) = 0, then we see that the above rule will not work,

So that-

Example: Find the P.I. Of (D + 2)(D – 1)2 y = e-2x + 2 sin h x

Sol.

Now we will evaluate each term separately-

And

Therefore-

Example: Solve (D – D’ – 2) (D – D’ – 3) z =

Sol.

The C.F. Will be given by-

Particular integral-

Therefore, the complete solution is-

z = C.F. + P.I. = e2x c1(y + x) + e3x c3(y+x) + 1/5 e3x – 2y

Case-2: when X = sin( ax + b) or cos (ax + b)

In case then the above rule fails.

Now-

And if

Similarly-

Example: Find the P.I. Of (D3 +1)y = cos (2x – 1)

Sol.

Example: Find the P.I. Of (D + 1) (D + D’ – 1)z = sin (x + 2y)

Sol.

Example: Find P.I. Of (D2 – 2D + 4)y = ex cos x

Sol.

Replace D by D+1

Put D2 = - 12 = - 1

Polynomials in x, eaxV(x) and xV(x).

Let us consider the equation-

Or in symbolic form-

(dn + k1dn – 1 + k2 dn – 2 + . . . + kn)y = X

So that-

P.I. = 1/(dn + k1dn – 1 + k2 dn – 2 + . . . + kn) X

Now-

Case-1: When X = eax, V is the function of x.

Case-2: When X is any other function of x.

Example: Find P.I. Of (D2 – 2D + 4) y = ex cos x

Sol.

Put

D2 = - 12 = -1

Working method to find the complete solution of an equation-

Example: Solve (D2 + 4D + 4)y = 3 sin x + 4 cos x

Sol.

Here first we will find the C.F.-

Its auxiliary equation will be-

D2 + 4D + 4 or (D + 2)2 = 0

Here we get-

D = -2, -2

C.F. = (c1 + c2x)e-2x

Now we will find P.I.-

Now the complete solution is-

Complete solution = C.F. + P.I.

= (c1 + c2 x)e-2x + sin x

Example: Solve-

Sol.

The given equation can be written as-

(D2 – 3D + 2) y = xe3x + sin 2x

Its auxiliary equation is-

D2 – 3D + 2 = 0 or (D – 2)(D – 1) = 0

We get-

D = 1, 2

So that the C.F. Will be-

C. F. = c1ex + c2e2x

Now we will find P.I.-

Therefore, the complete solution is-

Higher order linear differential equations with variable coefficients

The equation involving the derivatives of two or more dependent variables, which are the functions of a single independent variable, the equations involving their derivative is called simultaneous equation.

Such as-

We use method of elimination to solve these types of equations.

Example 1: Solve the following simultaneous differential equations-

....(2)

Solution:

Consider the given equations

....(2)

Consider eq(1),(2)

Dx+2y = et....(1)

Dx +2x =e-t....(2)

Eliminating ‘x’ from both the equations we get,

12 2Dx + 2y = 2et

D 2Dx +D2y = e-t

y = Ae2t + Be-2t +

Example 2: Solve the following simultaneous differential equations-

Given that x(0)=1 and y(0)= 0

Solution:

Consider the given equations,

Dy +2x = sin2t

Dx -2y = cos2t

By solving the above equations, we get,

(D2 +4)Y =0

X(0) = 1, y(0) = 0

A =0, B=-1

Example-3: Solve the following simultaneous differential equations-

It is given that x = 0 and y = 1 when t = 0.

Sol. Given equations can be written as-

Dx + 2y = - sin t ………. (1) and -2x + Dy = cos t ……… (2)

Eliminate x by multiplying (1) by 2 and (2) by D then add-

Here A.E =

So that C.F. =

And P.I. =

So that- …………. (3)

And ………….. (4)

Substitute (3) in (2), we get-

2x = Dy – cos t =

………… (5)

When t = 0, x = 0, y = 1. (3) and (5) gives-

Hence

Method of variation of parameters-

Consider a second order LDE with constant co-efficients given by

Then let the complimentary function yc is given by

yc = c1y1 + c2y2

Then the particular integral is

yp = uy1 + vy2

Where u and v are unknown and to be calculated using the formula

Example-1: Solve the following DE by using variation of parameters-

Sol. We can write the given equation in symbolic form as-

(D2 + 4)y = tan 2x

To find CF-

It’s A.E. Is (D2 + 4) = 0 or D = ± 2

So that CF is- y = c1 cos 2x + c2 sin 2x

To find PI-

Here y1 = cos 2x, y2 = sin 2x and X = tan 2x

Now W =

Thus PI = -

= - ¼ cos 2x log ( sec 2x + tan 2x)

So that the complete solution is-

y = c1 cos 2x + c2 sin 2x – ¼ cos 2x log (sec 2x + tan 2x)

Example-2: Solve the following by using the method of variation of parameters.

Sol. This can be written as-

(D2 – 2D + 1)y = ex log x

C.F.-

Auxiliary equation is- D2 – 2D + 1= 0 or (D – 1)2 so that D = 1, 1

So that the C.F. Will be- (c1 + c2x)ex

P.I.-

Here y1 = ex , y2 = xex and X = ex log x

Now W =

Thus PI = -

So that the complete solution is-

y = (c1 + c2x)ex + ¼ x2 ex(2 log x – 3)

Where, are constant is called Cauchy’s homogenous linear equation.

Put,

Example.

Ans. Putting,

AE is

CS = CF + PI

Example: Solve

Ans. Let,

AE is

y= CF + PI

Example: Solve

Ans. Let, so that z = log x

AE is

Legendre’s differential equation-

A differential equation of the form-

Where k is the constants and X is a function of x, is known as Legendre’s linear differential equation.

We can reduce such type of equations to linear equations with constant coefficient by the substitution as-

i.e.

Example: Solve

Sol. As we see that this is Legendre’s linear equation.

Now put

So that-

And

Then the equation becomes- D (D – 1)y+ Dy + y = 2 sin t

Its auxiliary equation is-

And particular integral-

P.I. =

Note -

Hence the solution is -

Key takeaways-

Where, are constant is called Cauchy’s homogenous linear equation.

2. A differential equation of the form-

Where k is the constants and X is a function of x, is known as Legendre’s linear differential equation.

The equation involving the derivatives of two or more dependent variables, which are the functions of a single independent variable, the equations involving their derivative is called simultaneous equation.

Such as-

We use method of elimination to solve these types of equations.

Example 1: Solve the following simultaneous differential equations-

....(2)

Solution:

Consider the given equations

....(2)

Consider eq(1),(2)

Dx+2y = et....(1)

Dx +2x =e-t....(2)

Eliminating ‘x’ from both the equations we get,

12 2Dx + 2y = 2et

D 2Dx +D2y = e-t

y = Ae2t + Be-2t +

Example 2: Solve the following simultaneous differential equations-

Given that x(0)=1 and y(0)= 0

Solution:

Consider the given equations,

Dy +2x = sin2t

Dx -2y = cos2t

By solving the above equations, we get,

(D2 +4)Y =0

X(0) = 1, y(0) = 0

A =0, B=-1

Example-3: Solve the following simultaneous differential equations-

It is given that x = 0 and y = 1 when t = 0.

Sol. Given equations can be written as-

Dx + 2y = - sin t ………. (1) and -2x + Dy = cos t ……… (2)

Eliminate x by multiplying (1) by 2 and (2) by D then add-

Here A.E =

So that C.F. =

And P.I. =

So that- …………. (3)

And ………….. (4)

Substitute (3) in (2), we get-

2x = Dy – cos t =

………… (5)

When t = 0, x = 0, y = 1. (3) and (5) gives-

Hence

The general form of linear differential equation of second order is

Where p and q are constants and R is a function of x or constant.

Differential Operator

D stands for operation of differential i.e.

Dy = dy/dx

D2y = d2y/dx2

1/D stands for the operator of integration.

1/D2 stands for operation of integration twice.

Thus,

(D2 + PD + q) = R

Note:- Complete solution = complementary function + Particular integral

i.e. y=CF + PI

Method for finding the CF

Step1:- In finding the CF right hand side of the given equation is replaced by zero.

Step 2:- Let y = C1 emx be the CF of

(D2 + PD + q) = 0 (1)

Putting the value of y dy/dx and d2y/dx2 in equation (1) we get

C1 emx (m2 + Pm + q) = 0

m2 + Pm + q = 0

It is called auxiliary equation.

Step 3:- Roots Real and Different

If m1 and m2 are the roots the CF is y = C1

If m1, m2 and m3 are the roots then CF y = C1

Step 4- Roots Real and Equal

If both the roots are m1 and m1 then CF is

y = (C1 + C2x)

If roots are m1, m1 and m1

y = (C1 + C2 x + C3 x2)

Example: Solve

(D2 + 8D + 15)y = 0

Ans. Given, (D2 + 8D + 15)y = 0

Here Auxiliary equation is

m2 – 8m + 15 = 9

m2 – 3m – 5m + 15 = 0

m(m – 3) – 5(m -3) = 0

(m – 3) (m – 5) = 0

m = 3, 5

y = C1e3x + C2 e5x

Solve:

Or,

(D2 – 6D + 9)y = 0

Ans. Auxiliary equation are m2 – 6m + 9 = 0

(m – 3)2 = 0 m = 3,3

CF = (C1 + C2x) e3x

Note: If roots are in complex form i.e. α = iβ

CF = eax [ Cos βx + Sin βx]

Solve: (D2 – 4D + 5)y = 0

Ans. Auxiliary equation are m2 – 4m + 8 = 0

m = -2 ± i

CF = e-2x[ C1 Cos x + C2 Sin x]

Solve. (D2 – 3D- 4)y = 0

Ans. Its auxiliary equation is

m2 – 3m – 4 = 0

m2 – 4m + m – 4 = 0

m(m – 4) + 1(m – 4) = 0

(m – 4)(m + 1) = 0

m = 4, - 1

Solution is y = C1 e4x + C2 e-x

Solve. D3 + D – D – 1)y = 0

Ans. The auxiliary equation is

m3 + m2 – m – 1 = 0

(m – 1)(m2 + 2m + 1) = 0

(m – 1)(m2 + 2m + 1) = 0

(m – 1)(m + 1)2 = 0

m = 1, -1, -1

Hence the solution is

y = C1ex + (C2 + C3x )e-x

Rules to find Particular Integral

Case 1:

If, f(a) = 0 then

If, f’(a) = 0 then

Solve:

Ans. Given, (D2 – 6D + 9)y = 5e3x

Auxiliary equation is m2 – 6m + 9 = 0

(m+3)2 = 0

m = -3, -3

CF = (C1 + C2x) e-3x

Put D = 3

CS = CF + PI

= (C1 + C2x) e-3x +

Case2:

Expand by the binomial theorem in ascending powers of D as far as the result of operation on xn is zero.

Solve. (D2 + 5D + 4)y = (3 – 2x)

Given, (D2 + 5D + 4)y = (3 – 2x)

For CF,

Auxiliary equation are m2 + 5m + 4 = 0

= m(m +4) + 1(m+4) = 0

= (m+1)(m+4) = 0

m = -1, -4

CF = C1e-x + C2e-4x

For PI

CS = CF + PI

= C1 e-x + C2e-4x +1/8 (11 – 4x)

Case 3:

Or,

Solve: (D2 + 4)y = Cos 2x

Ans. Auxiliary equation are m2 + 4 = 0

CF = A Cos 2x + B Sin 2x

PI =

Put , D2 = - a2 = -22 = -4

PI =

PI = x

= x/4 Sin 2x

CS = CF + PR

= A Cos 2x + B Sin 2x + x/4 Sin 2x

Case 4:

Solve. (D2 – 4D + 4)y = x3e2x

Ans. AE= m2 – 4m + 4 = 0

(m – 2)2 = 0 m = 2, 2

CF = (C1 + C2 X) e2x

Complete solution is y = (C1 + C2x) e2x + e2x

Solve (9D2 + 12D + 4)y = 6e-2x/3

Ans. The AE is 9m2 + 12m + 4 = 0

(3m + 2)2 = 0

(3m + 2)(3m + 2) = 0

m = -2/3, -2/3

CF = (C1 + C2 x) e-2x/3

Complete solution y= CF + PI

y = (C1 + C2x) e-2x/3 + x2/3

Solve. (D2 + 3D+2)y = 5

Ans. The AE is m2 + 3m + 2 = 0

m2 + m+ 2m + 2 = 0

m(m+1)+2(m+1) = 0

(m+1)(m+2) = 0

m = -1, -2

CF = C1 e-x + C2e-2x

Complete solution = CF + PI

C1e-x + C2e-2x + 5/2

Solve. (D2 – 6D +9)y = 2x2 – x +3

Ans. The AE is m2 – 6m + 9 = 0

(m – 3)2 = 0

m = 3, 3

CF = (C1 + C2x)e3x

Complete solutio0n is y= CF + PI

y – (C + C2x) e3x + 1/9 (2x2 + 5/3x + 11/3)

Find the PI of (D2 + 5D + 4) y = x2 + 7x + 9

Ans.

Solve

Ans. Given equation in symbolic form is

(D2 – 5D + 6)y = sin 3x

Its Auxiliary equation is m2 – 3m + 6 = 0

(m – 2)(m – 3) = 0

m = 2, 3

CF = C1e2x + C2e3x

Complete solution is y= CF + PI

y = C1e2x + C2e3x + 1/78 (5 cos 3x – sin 3x)

Solve. (D2 – 4)y = Cos2x

Ans. The AE is m2 – 4 = 0

m= ± 2

CF = C1 e2x + C2 e-2x

We know, cos 2x = 2 cos2x – 1

Cos2 x = (1+Cos 2x)/2

PI =

(since D2 -22 = -4)

= - 1/8 – 1/16 Cos 2x

Complete solution is y= CF + PI

y = C1e2x + C2e-2x – 1/8 – 1/16 Cos 2x

Solve. Find the PI of (D2 – 4D + 3)y = ex cos 2x

Ans.

Solve. (D3-7D-6) y=e2x (1+x)

Ans. The auxiliary equation i9s

m3 – 7m – 6 = 0

(m+1)(m2 –m – 6) = 0

(m+1)(m – 3)(m +2)=0

m = -1, 3, -2

CF = C1e-x + C2 e3x + C3 e-2x

Hence complete solution is y= CF + PI

y = C1e2x+ C2e3x+ C3e-2x – 1/12 e2x(17/12 + x)

Bessel equations

The Bessel equation is-

The solution of this equations will be-

The Bessel function is denoted by and defined as-

If we put n = 0 then Bessel function becomes-

Now if n = 1, then-

The graph of these two equations will be-

General solution of Bessel equation-

Example: Prove that-

Sol.

As we know that-

Jn(x) =

Now put n = 1/2 in equation (1), then we get-

J1/2(x) =

Hence proved.

Example: Prove that-

Sol.

Put n = -1/2 in equation (1) of the above question, we get-

J1/2(x) =

Recurrence formulae-

Formula-1:

Proof:

As we know that-

On differentiating with respect to x, we obtain-

Putting r – 1 = s

= nJn – x Jn+1

Formula-2:

Proof:

We have-

Differentiating w.r.t. x, we get-

Formula-3:

Proof:

We know that from formula first and second-

Now adding these two, we get-

Or

Formula-4:

Proof:

We know that-

XJ’n = nJn – x Jn-1

XJ’n = - nJn + xJn+1

On subtracting, we get-

0 = 2n Jn – x Jn+1 – x Jn -1

Formula-5:

Proof:

We know that-

Multiply this by we get-

x-n J’n = nx-n-1Jn – x-n Jn+1

I.e.

x-n J’n – n x-n-1 Jn = - x-n Jn+1

Or

Formula-6:

Proof:

We know that-

XJ’n = - n Jn + x Jn-1

Multiply by we get-

xnJ’n = -n xn-1Jn + xnJn-1

xnJ’n + n xn – 1Jn = xn Jn - 1

Or

Example: Show that-

By using recurrence relation.

Sol.

We know that-

The recurrence formula-

On differentiating, we get-

Now replace n by n -1 and n by n+1 in (1), we have-

2 J’n – 1 = Jn – 2 – Jn or J’n – 1 = ½ Jn – 2 – ½ Jn

2 J’n+1 = Jn – Jn+2 or J’n+1 = ½ Jn – ½ Jn+2

Put the values of and from the above equations in (2), we get-

2J’’n = ½ [ Jn-2 – Jn] – ½ [ Jn – Jn + 2]

4J’’n = Jn – 2 – Jn – Jn + Jn+2

4J’’n = Jn – 2 – 2 Jn + Jn+2

Example: Prove that-

Sol.

We know that- from recurrence formula

On integrating we get-

On taking n = 2 in (1), we get-

Again-

Put the value of from equation (2) and (3), we get-

By equation (1), when n = 1

Key takeaways-

- The Bessel function is denoted by and defined as-

2. General solution of Bessel equation-

Legendre’s equations

The Legendre’s equations is-

Now the solution of the given equation is the series of descending powers of x is-

Here is an arbitrary constant.

If n is a positive integer and

If f(x) is a continuous function having finite number of

Oscillations in the interval (0, a), then We can write

The above solution is

So that-

Here is called the Legendre’s function of first kind.

Note- Legendre’s equations of second kind is and can be defined as-

The general solution of Legendre’s equation is-

Here A and B are arbitrary constants.

Rodrigue’s formula-

Rodrigue’s formula can be defined as-

Legendre Polynomials-

We know that by Rodrigue formula-

If n = 0, then it becomes-

If n = 1,

If n = 2,

Now putting n =3, 4, 5……..n we get-

…………………………………..

Where N = n/2 if n is even and N = 1/2 (n-1) if n is odd.

Example: Express in terms of Legendre polynomials.

Sol.

By equating the coefficients of like powers of x, we get-

Put these values in equation (1), we get-

Example: Let be the Legendre’s polynomial of degree n, then show that for every function f(x) for which the n’th derivative is continuous-

Sol.

We know that-

On integrating by parts, we get-

Now integrate (n – 2) times by parts, we get-

Recurrence formulae for -

Formula-1:

NPn = (2n – 1)x Pn-1 – (n -1)Pn-2

Fromula-2:

XP’n – P’n – 1 = nPn

Formula-3:

P’n – xP’n-1 = nPn-1

Formula-4:

P’n+1 – P’n – 1 = (2n+1)Pn

Formula-5:

(x2-1)P’n = n[xPn – Pn-1]

Formula-6:

(x2 – 1)P’n =(n+1)(Pn+1 – xPn)

Generating function for

Prove that is the coefficient of in the expansion of in ascending powers of z.

Proof:

(1 – 2xz+z2)-1/2 = [1 – z(2x – z)]-1//2

Now coefficient of in

Coefficient of in

Coefficient of in

And so on.

Coefficient of in the expansion of equation (1)-

The coefficients of etc. in (1) are

Therefore-

(1 – 2xz + z2)-1/2 = P0(x) + zP1(x) + z2P2(x) + z3P3(x) + . . . + znPn(x) + …

Example: Show that-

P2n(0) = (-1)n

Sol.

We know that

Equating the coefficients of both sides, we have-

Orthogonality of Legendre polynomials-

Proof: is a solution of

…………………. (1)

And

is a solution of-

……………. (2)

Now multiply (1) by z and (2) by y and subtracting, we have-

Now integrate from -1 to +1, we get-

= 0

Example: Prove that-

By using Rodrigue formula for Legendre function.

On integrating by parts, we get-

Now integrating m – 2 times, we get-

Key takeaways-

- The Legendre’s equations is-

2. The general solution of Legendre’s equation is-

Here A and B are arbitrary constants.

3. Rodrigue’s formula can be defined as-

4. Orthogonality of Legendre polynomials-

References:

- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
- B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.

Unit - 3

Higher order linear differential equations with constant and variable coefficients

Higher order linear differential equations with constant coefficients

Second order linear homogeneous equations with constant coefficients

A homogeneous linear ordinary differential equation with constant coefficients is an ordinary differential equation in which coefficients are constants, all terms are linear, and the entire differential equation is equal to zero,

The form of second order linear differential equation with constant coefficients is,

Where a,b,c are the constants.

Let, aD²y+bDy+cy = f(x), where d² = d2/dx2 , D = d/dx

∅(D)y = f(x) , where ∅(D)y = aD²y+bDy+cy

Here first we solve, ∅(D)y = 0, which is called complementary function(C.F)

Then we find particular integral (P.I)

P.I. = f(x)

General solution = C.F. +P.I.

Let’s do an example to understand the concept,

Example1: Solve (4D² +4D -3)y = e2x

Solution: Auxiliary equation is 4m² +4m – 3 = 0

We get, (2m+3) (2m – 1) = 0

m = -3/2, 1/2

Complementary function: CF is A e-3x/2+ B e1 x/2

Now we will find particular integral,

P.I. = f(x)

General solution is y = CF + PI

= A e-3x/2+ B e1 x/2 + 1/21 . e2x

Non-Homogeneous equations of higher order with constant coefficient of R.H.S. Terms of the type eax, sin ax, cos ax

Linear differential equation is those in which the independent variable and its derivative occur only in the first degree and are not multiplied together.

Thus, the general linear differential equation of the n’th order is of the form

Where p1, p2, . . ., pn and X are function of x.

Linear differential equation with constant co-efficient are of the form-

Where k1, k2, . . . , kn are constants.

Rules to find the complementary function-

To solve the equation-

This can be written as in symbolic form-

(dn + k1dn – 1 + k2dn – 2 + . . . + kn)y = 0 ………….(2)

Or-

(dn + k1dn – 1 + k2dn – 2 + . . . + kn) = 0

It is called the auxiliary equation.

Let m1, m2, . . . , mn be the roots-

Case-1: If all the roots are real and distinct, then equation (2) becomes,

(D – m1)(D – m2) . . . (D – mn)y = 0

Now this equation will be satisfied by the solution of (D – mn)y = 0

This is a Leibnitz’s linear and I.F. =

Its solution is

The complete solution will be-

Case-2: If two roots are equal m1 = m2

Then complete solution is given by-

y = (c1x2 + c2x + c3)

Case-3: If one pair of roots be imaginary, i.e. m1 = α + iβ, m2 = α - iβ then the complete solution is-

y = eax (c1 cos β x + c2 sin β x) + c3

Where C1 = c1 + c2 and C2 = i(c1 – c2)

Case-4: If two points of imaginary roots be equal-

m1 = m2 = α + iβ, m3= m4 = α - iβ

Then the complete solution is-

y = eax[(c1 x + c2) cos β x + (c3x + c4) sin β x)] + . . . + cn

Example-Solve (D2 + 6D + 9) = 0

Sol.

Its auxiliary equation is-

D2 + 6D + 9 = 0

(D + 3)2 = 0

Where-

D = -3, -3

Therefore, the complete solution is-

x = (c1 + c2t)e-3t

Inverse operator-

is that function of x, not containing arbitrary constants which when operated upon gives X.

So that satisfies the equation f(D)y = X and is, therefore, its P.I.

f(D) and 1/f(D) are inverse operator.

Note-

1.

2.

Rules for finding the particular integral-

Let us consider the equation-

Or in symbolic form-

(dn + k1dn – 1 + k2 dn – 2 + . . . + kn)y = X

So that-

P.I. = 1/(dn + k1dn – 1 + k2 dn – 2 + . . . + kn) X

Now-

Case-1: When X = eax

In case f(a) = 0, then we see that the above rule will not work,

So that-

Example: Find the P.I. Of (D + 2)(D – 1)2 y = e-2x + 2 sin h x

Sol.

Now we will evaluate each term separately-

And

Therefore-

Example: Solve (D – D’ – 2) (D – D’ – 3) z =

Sol.

The C.F. Will be given by-

Particular integral-

Therefore, the complete solution is-

z = C.F. + P.I. = e2x c1(y + x) + e3x c3(y+x) + 1/5 e3x – 2y

Case-2: when X = sin( ax + b) or cos (ax + b)

In case then the above rule fails.

Now-

And if

Similarly-

Example: Find the P.I. Of (D3 +1)y = cos (2x – 1)

Sol.

Example: Find the P.I. Of (D + 1) (D + D’ – 1)z = sin (x + 2y)

Sol.

Example: Find P.I. Of (D2 – 2D + 4)y = ex cos x

Sol.

Replace D by D+1

Put D2 = - 12 = - 1

Polynomials in x, eaxV(x) and xV(x).

Let us consider the equation-

Or in symbolic form-

(dn + k1dn – 1 + k2 dn – 2 + . . . + kn)y = X

So that-

P.I. = 1/(dn + k1dn – 1 + k2 dn – 2 + . . . + kn) X

Now-

Case-1: When X = eax, V is the function of x.

Case-2: When X is any other function of x.

Example: Find P.I. Of (D2 – 2D + 4) y = ex cos x

Sol.

Put

D2 = - 12 = -1

Working method to find the complete solution of an equation-

Example: Solve (D2 + 4D + 4)y = 3 sin x + 4 cos x

Sol.

Here first we will find the C.F.-

Its auxiliary equation will be-

D2 + 4D + 4 or (D + 2)2 = 0

Here we get-

D = -2, -2

C.F. = (c1 + c2x)e-2x

Now we will find P.I.-

Now the complete solution is-

Complete solution = C.F. + P.I.

= (c1 + c2 x)e-2x + sin x

Example: Solve-

Sol.

The given equation can be written as-

(D2 – 3D + 2) y = xe3x + sin 2x

Its auxiliary equation is-

D2 – 3D + 2 = 0 or (D – 2)(D – 1) = 0

We get-

D = 1, 2

So that the C.F. Will be-

C. F. = c1ex + c2e2x

Now we will find P.I.-

Therefore, the complete solution is-

Higher order linear differential equations with variable coefficients

The equation involving the derivatives of two or more dependent variables, which are the functions of a single independent variable, the equations involving their derivative is called simultaneous equation.

Such as-

We use method of elimination to solve these types of equations.

Example 1: Solve the following simultaneous differential equations-

....(2)

Solution:

Consider the given equations

....(2)

Consider eq(1),(2)

Dx+2y = et....(1)

Dx +2x =e-t....(2)

Eliminating ‘x’ from both the equations we get,

12 2Dx + 2y = 2et

D 2Dx +D2y = e-t

y = Ae2t + Be-2t +

Example 2: Solve the following simultaneous differential equations-

Given that x(0)=1 and y(0)= 0

Solution:

Consider the given equations,

Dy +2x = sin2t

Dx -2y = cos2t

By solving the above equations, we get,

(D2 +4)Y =0

X(0) = 1, y(0) = 0

A =0, B=-1

Example-3: Solve the following simultaneous differential equations-

It is given that x = 0 and y = 1 when t = 0.

Sol. Given equations can be written as-

Dx + 2y = - sin t ………. (1) and -2x + Dy = cos t ……… (2)

Eliminate x by multiplying (1) by 2 and (2) by D then add-

Here A.E =

So that C.F. =

And P.I. =

So that- …………. (3)

And ………….. (4)

Substitute (3) in (2), we get-

2x = Dy – cos t =

………… (5)

When t = 0, x = 0, y = 1. (3) and (5) gives-

Hence

Method of variation of parameters-

Consider a second order LDE with constant co-efficients given by

Then let the complimentary function yc is given by

yc = c1y1 + c2y2

Then the particular integral is

yp = uy1 + vy2

Where u and v are unknown and to be calculated using the formula

Example-1: Solve the following DE by using variation of parameters-

Sol. We can write the given equation in symbolic form as-

(D2 + 4)y = tan 2x

To find CF-

It’s A.E. Is (D2 + 4) = 0 or D = ± 2

So that CF is- y = c1 cos 2x + c2 sin 2x

To find PI-

Here y1 = cos 2x, y2 = sin 2x and X = tan 2x

Now W =

Thus PI = -

= - ¼ cos 2x log ( sec 2x + tan 2x)

So that the complete solution is-

y = c1 cos 2x + c2 sin 2x – ¼ cos 2x log (sec 2x + tan 2x)

Example-2: Solve the following by using the method of variation of parameters.

Sol. This can be written as-

(D2 – 2D + 1)y = ex log x

C.F.-

Auxiliary equation is- D2 – 2D + 1= 0 or (D – 1)2 so that D = 1, 1

So that the C.F. Will be- (c1 + c2x)ex

P.I.-

Here y1 = ex , y2 = xex and X = ex log x

Now W =

Thus PI = -

So that the complete solution is-

y = (c1 + c2x)ex + ¼ x2 ex(2 log x – 3)

Where, are constant is called Cauchy’s homogenous linear equation.

Put,

Example.

Ans. Putting,

AE is

CS = CF + PI

Example: Solve

Ans. Let,

AE is

y= CF + PI

Example: Solve

Ans. Let, so that z = log x

AE is

Legendre’s differential equation-

A differential equation of the form-

Where k is the constants and X is a function of x, is known as Legendre’s linear differential equation.

We can reduce such type of equations to linear equations with constant coefficient by the substitution as-

i.e.

Example: Solve

Sol. As we see that this is Legendre’s linear equation.

Now put

So that-

And

Then the equation becomes- D (D – 1)y+ Dy + y = 2 sin t

Its auxiliary equation is-

And particular integral-

P.I. =

Note -

Hence the solution is -

Key takeaways-

Where, are constant is called Cauchy’s homogenous linear equation.

2. A differential equation of the form-

Such as-

We use method of elimination to solve these types of equations.

Example 1: Solve the following simultaneous differential equations-

....(2)

Solution:

Consider the given equations

....(2)

Consider eq(1),(2)

Dx+2y = et....(1)

Dx +2x =e-t....(2)

Eliminating ‘x’ from both the equations we get,

12 2Dx + 2y = 2et

D 2Dx +D2y = e-t

y = Ae2t + Be-2t +

Example 2: Solve the following simultaneous differential equations-

Given that x(0)=1 and y(0)= 0

Solution:

Consider the given equations,

Dy +2x = sin2t

Dx -2y = cos2t

By solving the above equations, we get,

(D2 +4)Y =0

X(0) = 1, y(0) = 0

A =0, B=-1

Example-3: Solve the following simultaneous differential equations-

It is given that x = 0 and y = 1 when t = 0.

Sol. Given equations can be written as-

Dx + 2y = - sin t ………. (1) and -2x + Dy = cos t ……… (2)

Eliminate x by multiplying (1) by 2 and (2) by D then add-

Here A.E =

So that C.F. =

And P.I. =

So that- …………. (3)

And ………….. (4)

Substitute (3) in (2), we get-

2x = Dy – cos t =

………… (5)

When t = 0, x = 0, y = 1. (3) and (5) gives-

Hence

The general form of linear differential equation of second order is

Where p and q are constants and R is a function of x or constant.

Differential Operator

D stands for operation of differential i.e.

Dy = dy/dx

D2y = d2y/dx2

1/D stands for the operator of integration.

1/D2 stands for operation of integration twice.

Thus,

(D2 + PD + q) = R

Note:- Complete solution = complementary function + Particular integral

i.e. y=CF + PI

Method for finding the CF

Step1:- In finding the CF right hand side of the given equation is replaced by zero.

Step 2:- Let y = C1 emx be the CF of

(D2 + PD + q) = 0 (1)

Putting the value of y dy/dx and d2y/dx2 in equation (1) we get

C1 emx (m2 + Pm + q) = 0

m2 + Pm + q = 0

It is called auxiliary equation.

Step 3:- Roots Real and Different

If m1 and m2 are the roots the CF is y = C1

If m1, m2 and m3 are the roots then CF y = C1

Step 4- Roots Real and Equal

If both the roots are m1 and m1 then CF is

y = (C1 + C2x)

If roots are m1, m1 and m1

y = (C1 + C2 x + C3 x2)

Example: Solve

(D2 + 8D + 15)y = 0

Ans. Given, (D2 + 8D + 15)y = 0

Here Auxiliary equation is

m2 – 8m + 15 = 9

m2 – 3m – 5m + 15 = 0

m(m – 3) – 5(m -3) = 0

(m – 3) (m – 5) = 0

m = 3, 5

y = C1e3x + C2 e5x

Solve:

Or,

(D2 – 6D + 9)y = 0

Ans. Auxiliary equation are m2 – 6m + 9 = 0

(m – 3)2 = 0 m = 3,3

CF = (C1 + C2x) e3x

Note: If roots are in complex form i.e. α = iβ

CF = eax [ Cos βx + Sin βx]

Solve: (D2 – 4D + 5)y = 0

Ans. Auxiliary equation are m2 – 4m + 8 = 0

m = -2 ± i

CF = e-2x[ C1 Cos x + C2 Sin x]

Solve. (D2 – 3D- 4)y = 0

Ans. Its auxiliary equation is

m2 – 3m – 4 = 0

m2 – 4m + m – 4 = 0

m(m – 4) + 1(m – 4) = 0

(m – 4)(m + 1) = 0

m = 4, - 1

Solution is y = C1 e4x + C2 e-x

Solve. D3 + D – D – 1)y = 0

Ans. The auxiliary equation is

m3 + m2 – m – 1 = 0

(m – 1)(m2 + 2m + 1) = 0

(m – 1)(m2 + 2m + 1) = 0

(m – 1)(m + 1)2 = 0

m = 1, -1, -1

Hence the solution is

y = C1ex + (C2 + C3x )e-x

Rules to find Particular Integral

Case 1:

If, f(a) = 0 then

If, f’(a) = 0 then

Solve:

Ans. Given, (D2 – 6D + 9)y = 5e3x

Auxiliary equation is m2 – 6m + 9 = 0

(m+3)2 = 0

m = -3, -3

CF = (C1 + C2x) e-3x

Put D = 3

CS = CF + PI

= (C1 + C2x) e-3x +

Case2:

Expand by the binomial theorem in ascending powers of D as far as the result of operation on xn is zero.

Solve. (D2 + 5D + 4)y = (3 – 2x)

Given, (D2 + 5D + 4)y = (3 – 2x)

For CF,

Auxiliary equation are m2 + 5m + 4 = 0

= m(m +4) + 1(m+4) = 0

= (m+1)(m+4) = 0

m = -1, -4

CF = C1e-x + C2e-4x

For PI

CS = CF + PI

= C1 e-x + C2e-4x +1/8 (11 – 4x)

Case 3:

Or,

Solve: (D2 + 4)y = Cos 2x

Ans. Auxiliary equation are m2 + 4 = 0

CF = A Cos 2x + B Sin 2x

PI =

Put , D2 = - a2 = -22 = -4

PI =

PI = x

= x/4 Sin 2x

CS = CF + PR

= A Cos 2x + B Sin 2x + x/4 Sin 2x

Case 4:

Solve. (D2 – 4D + 4)y = x3e2x

Ans. AE= m2 – 4m + 4 = 0

(m – 2)2 = 0 m = 2, 2

CF = (C1 + C2 X) e2x

Complete solution is y = (C1 + C2x) e2x + e2x

Solve (9D2 + 12D + 4)y = 6e-2x/3

Ans. The AE is 9m2 + 12m + 4 = 0

(3m + 2)2 = 0

(3m + 2)(3m + 2) = 0

m = -2/3, -2/3

CF = (C1 + C2 x) e-2x/3

Complete solution y= CF + PI

y = (C1 + C2x) e-2x/3 + x2/3

Solve. (D2 + 3D+2)y = 5

Ans. The AE is m2 + 3m + 2 = 0

m2 + m+ 2m + 2 = 0

m(m+1)+2(m+1) = 0

(m+1)(m+2) = 0

m = -1, -2

CF = C1 e-x + C2e-2x

Complete solution = CF + PI

C1e-x + C2e-2x + 5/2

Solve. (D2 – 6D +9)y = 2x2 – x +3

Ans. The AE is m2 – 6m + 9 = 0

(m – 3)2 = 0

m = 3, 3

CF = (C1 + C2x)e3x

Complete solutio0n is y= CF + PI

y – (C + C2x) e3x + 1/9 (2x2 + 5/3x + 11/3)

Find the PI of (D2 + 5D + 4) y = x2 + 7x + 9

Ans.

Solve

Ans. Given equation in symbolic form is

(D2 – 5D + 6)y = sin 3x

Its Auxiliary equation is m2 – 3m + 6 = 0

(m – 2)(m – 3) = 0

m = 2, 3

CF = C1e2x + C2e3x

Complete solution is y= CF + PI

y = C1e2x + C2e3x + 1/78 (5 cos 3x – sin 3x)

Solve. (D2 – 4)y = Cos2x

Ans. The AE is m2 – 4 = 0

m= ± 2

CF = C1 e2x + C2 e-2x

We know, cos 2x = 2 cos2x – 1

Cos2 x = (1+Cos 2x)/2

PI =

(since D2 -22 = -4)

= - 1/8 – 1/16 Cos 2x

Complete solution is y= CF + PI

y = C1e2x + C2e-2x – 1/8 – 1/16 Cos 2x

Solve. Find the PI of (D2 – 4D + 3)y = ex cos 2x

Ans.

Solve. (D3-7D-6) y=e2x (1+x)

Ans. The auxiliary equation i9s

m3 – 7m – 6 = 0

(m+1)(m2 –m – 6) = 0

(m+1)(m – 3)(m +2)=0

m = -1, 3, -2

CF = C1e-x + C2 e3x + C3 e-2x

Hence complete solution is y= CF + PI

y = C1e2x+ C2e3x+ C3e-2x – 1/12 e2x(17/12 + x)

Bessel equations

The Bessel equation is-

The solution of this equations will be-

The Bessel function is denoted by and defined as-

If we put n = 0 then Bessel function becomes-

Now if n = 1, then-

The graph of these two equations will be-

General solution of Bessel equation-

Example: Prove that-

Sol.

As we know that-

Jn(x) =

Now put n = 1/2 in equation (1), then we get-

J1/2(x) =

Hence proved.

Example: Prove that-

Sol.

Put n = -1/2 in equation (1) of the above question, we get-

J1/2(x) =

Recurrence formulae-

Formula-1:

Proof:

As we know that-

On differentiating with respect to x, we obtain-

Putting r – 1 = s

= nJn – x Jn+1

Formula-2:

Proof:

We have-

Differentiating w.r.t. x, we get-

Formula-3:

Proof:

We know that from formula first and second-

Now adding these two, we get-

Or

Formula-4:

Proof:

We know that-

XJ’n = nJn – x Jn-1

XJ’n = - nJn + xJn+1

On subtracting, we get-

0 = 2n Jn – x Jn+1 – x Jn -1

Formula-5:

Proof:

We know that-

Multiply this by we get-

x-n J’n = nx-n-1Jn – x-n Jn+1

I.e.

x-n J’n – n x-n-1 Jn = - x-n Jn+1

Or

Formula-6:

Proof:

We know that-

XJ’n = - n Jn + x Jn-1

Multiply by we get-

xnJ’n = -n xn-1Jn + xnJn-1

xnJ’n + n xn – 1Jn = xn Jn - 1

Or

Example: Show that-

By using recurrence relation.

Sol.

We know that-

The recurrence formula-

On differentiating, we get-

Now replace n by n -1 and n by n+1 in (1), we have-

2 J’n – 1 = Jn – 2 – Jn or J’n – 1 = ½ Jn – 2 – ½ Jn

2 J’n+1 = Jn – Jn+2 or J’n+1 = ½ Jn – ½ Jn+2

Put the values of and from the above equations in (2), we get-

2J’’n = ½ [ Jn-2 – Jn] – ½ [ Jn – Jn + 2]

4J’’n = Jn – 2 – Jn – Jn + Jn+2

4J’’n = Jn – 2 – 2 Jn + Jn+2

Example: Prove that-

Sol.

We know that- from recurrence formula

On integrating we get-

On taking n = 2 in (1), we get-

Again-

Put the value of from equation (2) and (3), we get-

By equation (1), when n = 1

Key takeaways-

- The Bessel function is denoted by and defined as-

2. General solution of Bessel equation-

Legendre’s equations

The Legendre’s equations is-

Now the solution of the given equation is the series of descending powers of x is-

Here is an arbitrary constant.

If n is a positive integer and

If f(x) is a continuous function having finite number of

Oscillations in the interval (0, a), then We can write

The above solution is

So that-

Here is called the Legendre’s function of first kind.

Note- Legendre’s equations of second kind is and can be defined as-

The general solution of Legendre’s equation is-

Here A and B are arbitrary constants.

Rodrigue’s formula-

Rodrigue’s formula can be defined as-

Legendre Polynomials-

We know that by Rodrigue formula-

If n = 0, then it becomes-

If n = 1,

If n = 2,

Now putting n =3, 4, 5……..n we get-

…………………………………..

Where N = n/2 if n is even and N = 1/2 (n-1) if n is odd.

Example: Express in terms of Legendre polynomials.

Sol.

By equating the coefficients of like powers of x, we get-

Put these values in equation (1), we get-

Example: Let be the Legendre’s polynomial of degree n, then show that for every function f(x) for which the n’th derivative is continuous-

Sol.

We know that-

On integrating by parts, we get-

Now integrate (n – 2) times by parts, we get-

Recurrence formulae for -

Formula-1:

NPn = (2n – 1)x Pn-1 – (n -1)Pn-2

Fromula-2:

XP’n – P’n – 1 = nPn

Formula-3:

P’n – xP’n-1 = nPn-1

Formula-4:

P’n+1 – P’n – 1 = (2n+1)Pn

Formula-5:

(x2-1)P’n = n[xPn – Pn-1]

Formula-6:

(x2 – 1)P’n =(n+1)(Pn+1 – xPn)

Generating function for

Prove that is the coefficient of in the expansion of in ascending powers of z.

Proof:

(1 – 2xz+z2)-1/2 = [1 – z(2x – z)]-1//2

Now coefficient of in

Coefficient of in

Coefficient of in

And so on.

Coefficient of in the expansion of equation (1)-

The coefficients of etc. in (1) are

Therefore-

(1 – 2xz + z2)-1/2 = P0(x) + zP1(x) + z2P2(x) + z3P3(x) + . . . + znPn(x) + …

Example: Show that-

P2n(0) = (-1)n

Sol.

We know that

Equating the coefficients of both sides, we have-

Orthogonality of Legendre polynomials-

Proof: is a solution of

…………………. (1)

And

is a solution of-

……………. (2)

Now multiply (1) by z and (2) by y and subtracting, we have-

Now integrate from -1 to +1, we get-

= 0

Example: Prove that-

By using Rodrigue formula for Legendre function.

On integrating by parts, we get-

Now integrating m – 2 times, we get-

Key takeaways-

- The Legendre’s equations is-

2. The general solution of Legendre’s equation is-

Here A and B are arbitrary constants.

3. Rodrigue’s formula can be defined as-

4. Orthogonality of Legendre polynomials-

References:

- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
- B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.

Unit - 3

Higher order linear differential equations with constant and variable coefficients

Unit - 3

Higher order linear differential equations with constant and variable coefficients

Unit - 3

Higher order linear differential equations with constant and variable coefficients

Unit - 3

Higher order linear differential equations with constant and variable coefficients

Unit - 3

Higher order linear differential equations with constant and variable coefficients

Higher order linear differential equations with constant coefficients

Second order linear homogeneous equations with constant coefficients

A homogeneous linear ordinary differential equation with constant coefficients is an ordinary differential equation in which coefficients are constants, all terms are linear, and the entire differential equation is equal to zero,

The form of second order linear differential equation with constant coefficients is,

Where a,b,c are the constants.

Let, aD²y+bDy+cy = f(x), where d² = d2/dx2 , D = d/dx

∅(D)y = f(x) , where ∅(D)y = aD²y+bDy+cy

Here first we solve, ∅(D)y = 0, which is called complementary function(C.F)

Then we find particular integral (P.I)

P.I. = f(x)

General solution = C.F. +P.I.

Let’s do an example to understand the concept,

Example1: Solve (4D² +4D -3)y = e2x

Solution: Auxiliary equation is 4m² +4m – 3 = 0

We get, (2m+3) (2m – 1) = 0

m = -3/2, 1/2

Complementary function: CF is A e-3x/2+ B e1 x/2

Now we will find particular integral,

P.I. = f(x)

General solution is y = CF + PI

= A e-3x/2+ B e1 x/2 + 1/21 . e2x

Non-Homogeneous equations of higher order with constant coefficient of R.H.S. Terms of the type eax, sin ax, cos ax

Linear differential equation is those in which the independent variable and its derivative occur only in the first degree and are not multiplied together.

Thus, the general linear differential equation of the n’th order is of the form

Where p1, p2, . . ., pn and X are function of x.

Linear differential equation with constant co-efficient are of the form-

Where k1, k2, . . . , kn are constants.

Rules to find the complementary function-

To solve the equation-

This can be written as in symbolic form-

(dn + k1dn – 1 + k2dn – 2 + . . . + kn)y = 0 ………….(2)

Or-

(dn + k1dn – 1 + k2dn – 2 + . . . + kn) = 0

It is called the auxiliary equation.

Let m1, m2, . . . , mn be the roots-

Case-1: If all the roots are real and distinct, then equation (2) becomes,

(D – m1)(D – m2) . . . (D – mn)y = 0

Now this equation will be satisfied by the solution of (D – mn)y = 0

This is a Leibnitz’s linear and I.F. =

Its solution is

The complete solution will be-

Case-2: If two roots are equal m1 = m2

Then complete solution is given by-

y = (c1x2 + c2x + c3)

Case-3: If one pair of roots be imaginary, i.e. m1 = α + iβ, m2 = α - iβ then the complete solution is-

y = eax (c1 cos β x + c2 sin β x) + c3

Where C1 = c1 + c2 and C2 = i(c1 – c2)

Case-4: If two points of imaginary roots be equal-

m1 = m2 = α + iβ, m3= m4 = α - iβ

Then the complete solution is-

y = eax[(c1 x + c2) cos β x + (c3x + c4) sin β x)] + . . . + cn

Example-Solve (D2 + 6D + 9) = 0

Sol.

Its auxiliary equation is-

D2 + 6D + 9 = 0

(D + 3)2 = 0

Where-

D = -3, -3

Therefore, the complete solution is-

x = (c1 + c2t)e-3t

Inverse operator-

is that function of x, not containing arbitrary constants which when operated upon gives X.

So that satisfies the equation f(D)y = X and is, therefore, its P.I.

f(D) and 1/f(D) are inverse operator.

Note-

1.

2.

Rules for finding the particular integral-

Let us consider the equation-

Or in symbolic form-

(dn + k1dn – 1 + k2 dn – 2 + . . . + kn)y = X

So that-

P.I. = 1/(dn + k1dn – 1 + k2 dn – 2 + . . . + kn) X

Now-

Case-1: When X = eax

In case f(a) = 0, then we see that the above rule will not work,

So that-

Example: Find the P.I. Of (D + 2)(D – 1)2 y = e-2x + 2 sin h x

Sol.

Now we will evaluate each term separately-

And

Therefore-

Example: Solve (D – D’ – 2) (D – D’ – 3) z =

Sol.

The C.F. Will be given by-

Particular integral-

Therefore, the complete solution is-

z = C.F. + P.I. = e2x c1(y + x) + e3x c3(y+x) + 1/5 e3x – 2y

Case-2: when X = sin( ax + b) or cos (ax + b)

In case then the above rule fails.

Now-

And if

Similarly-

Example: Find the P.I. Of (D3 +1)y = cos (2x – 1)

Sol.

Example: Find the P.I. Of (D + 1) (D + D’ – 1)z = sin (x + 2y)

Sol.

Example: Find P.I. Of (D2 – 2D + 4)y = ex cos x

Sol.

Replace D by D+1

Put D2 = - 12 = - 1

Polynomials in x, eaxV(x) and xV(x).

Let us consider the equation-

Or in symbolic form-

(dn + k1dn – 1 + k2 dn – 2 + . . . + kn)y = X

So that-

P.I. = 1/(dn + k1dn – 1 + k2 dn – 2 + . . . + kn) X

Now-

Case-1: When X = eax, V is the function of x.

Case-2: When X is any other function of x.

Example: Find P.I. Of (D2 – 2D + 4) y = ex cos x

Sol.

Put

D2 = - 12 = -1

Working method to find the complete solution of an equation-

Example: Solve (D2 + 4D + 4)y = 3 sin x + 4 cos x

Sol.

Here first we will find the C.F.-

Its auxiliary equation will be-

D2 + 4D + 4 or (D + 2)2 = 0

Here we get-

D = -2, -2

C.F. = (c1 + c2x)e-2x

Now we will find P.I.-

Now the complete solution is-

Complete solution = C.F. + P.I.

= (c1 + c2 x)e-2x + sin x

Example: Solve-

Sol.

The given equation can be written as-

(D2 – 3D + 2) y = xe3x + sin 2x

Its auxiliary equation is-

D2 – 3D + 2 = 0 or (D – 2)(D – 1) = 0

We get-

D = 1, 2

So that the C.F. Will be-

C. F. = c1ex + c2e2x

Now we will find P.I.-

Therefore, the complete solution is-

Higher order linear differential equations with variable coefficients

Such as-

We use method of elimination to solve these types of equations.

Example 1: Solve the following simultaneous differential equations-

....(2)

Solution:

Consider the given equations

....(2)

Consider eq(1),(2)

Dx+2y = et....(1)

Dx +2x =e-t....(2)

Eliminating ‘x’ from both the equations we get,

12 2Dx + 2y = 2et

D 2Dx +D2y = e-t

y = Ae2t + Be-2t +

Example 2: Solve the following simultaneous differential equations-

Given that x(0)=1 and y(0)= 0

Solution:

Consider the given equations,

Dy +2x = sin2t

Dx -2y = cos2t

By solving the above equations, we get,

(D2 +4)Y =0

X(0) = 1, y(0) = 0

A =0, B=-1

Example-3: Solve the following simultaneous differential equations-

It is given that x = 0 and y = 1 when t = 0.

Sol. Given equations can be written as-

Dx + 2y = - sin t ………. (1) and -2x + Dy = cos t ……… (2)

Eliminate x by multiplying (1) by 2 and (2) by D then add-

Here A.E =

So that C.F. =

And P.I. =

So that- …………. (3)

And ………….. (4)

Substitute (3) in (2), we get-

2x = Dy – cos t =

………… (5)

When t = 0, x = 0, y = 1. (3) and (5) gives-

Hence

Method of variation of parameters-

Consider a second order LDE with constant co-efficients given by

Then let the complimentary function yc is given by

yc = c1y1 + c2y2

Then the particular integral is

yp = uy1 + vy2

Where u and v are unknown and to be calculated using the formula

Example-1: Solve the following DE by using variation of parameters-

Sol. We can write the given equation in symbolic form as-

(D2 + 4)y = tan 2x

To find CF-

It’s A.E. Is (D2 + 4) = 0 or D = ± 2

So that CF is- y = c1 cos 2x + c2 sin 2x

To find PI-

Here y1 = cos 2x, y2 = sin 2x and X = tan 2x

Now W =

Thus PI = -

= - ¼ cos 2x log ( sec 2x + tan 2x)

So that the complete solution is-

y = c1 cos 2x + c2 sin 2x – ¼ cos 2x log (sec 2x + tan 2x)

Example-2: Solve the following by using the method of variation of parameters.

Sol. This can be written as-

(D2 – 2D + 1)y = ex log x

C.F.-

Auxiliary equation is- D2 – 2D + 1= 0 or (D – 1)2 so that D = 1, 1

So that the C.F. Will be- (c1 + c2x)ex

P.I.-

Here y1 = ex , y2 = xex and X = ex log x

Now W =

Thus PI = -

So that the complete solution is-

y = (c1 + c2x)ex + ¼ x2 ex(2 log x – 3)

Where, are constant is called Cauchy’s homogenous linear equation.

Put,

Example.

Ans. Putting,

AE is

CS = CF + PI

Example: Solve

Ans. Let,

AE is

y= CF + PI

Example: Solve

Ans. Let, so that z = log x

AE is

Legendre’s differential equation-

A differential equation of the form-

We can reduce such type of equations to linear equations with constant coefficient by the substitution as-

i.e.

Example: Solve

Sol. As we see that this is Legendre’s linear equation.

Now put

So that-

And

Then the equation becomes- D (D – 1)y+ Dy + y = 2 sin t

Its auxiliary equation is-

And particular integral-

P.I. =

Note -

Hence the solution is -

Key takeaways-

Where, are constant is called Cauchy’s homogenous linear equation.

2. A differential equation of the form-

Such as-

We use method of elimination to solve these types of equations.

Example 1: Solve the following simultaneous differential equations-

....(2)

Solution:

Consider the given equations

....(2)

Consider eq(1),(2)

Dx+2y = et....(1)

Dx +2x =e-t....(2)

Eliminating ‘x’ from both the equations we get,

12 2Dx + 2y = 2et

D 2Dx +D2y = e-t

y = Ae2t + Be-2t +

Example 2: Solve the following simultaneous differential equations-

Given that x(0)=1 and y(0)= 0

Solution:

Consider the given equations,

Dy +2x = sin2t

Dx -2y = cos2t

By solving the above equations, we get,

(D2 +4)Y =0

X(0) = 1, y(0) = 0

A =0, B=-1

Example-3: Solve the following simultaneous differential equations-

It is given that x = 0 and y = 1 when t = 0.

Sol. Given equations can be written as-

Dx + 2y = - sin t ………. (1) and -2x + Dy = cos t ……… (2)

Eliminate x by multiplying (1) by 2 and (2) by D then add-

Here A.E =

So that C.F. =

And P.I. =

So that- …………. (3)

And ………….. (4)

Substitute (3) in (2), we get-

2x = Dy – cos t =

………… (5)

When t = 0, x = 0, y = 1. (3) and (5) gives-

Hence

The general form of linear differential equation of second order is

Where p and q are constants and R is a function of x or constant.

Differential Operator

D stands for operation of differential i.e.

Dy = dy/dx

D2y = d2y/dx2

1/D stands for the operator of integration.

1/D2 stands for operation of integration twice.

Thus,

(D2 + PD + q) = R

Note:- Complete solution = complementary function + Particular integral

i.e. y=CF + PI

Method for finding the CF

Step1:- In finding the CF right hand side of the given equation is replaced by zero.

Step 2:- Let y = C1 emx be the CF of

(D2 + PD + q) = 0 (1)

Putting the value of y dy/dx and d2y/dx2 in equation (1) we get

C1 emx (m2 + Pm + q) = 0

m2 + Pm + q = 0

It is called auxiliary equation.

Step 3:- Roots Real and Different

If m1 and m2 are the roots the CF is y = C1

If m1, m2 and m3 are the roots then CF y = C1

Step 4- Roots Real and Equal

If both the roots are m1 and m1 then CF is

y = (C1 + C2x)

If roots are m1, m1 and m1

y = (C1 + C2 x + C3 x2)

Example: Solve

(D2 + 8D + 15)y = 0

Ans. Given, (D2 + 8D + 15)y = 0

Here Auxiliary equation is

m2 – 8m + 15 = 9

m2 – 3m – 5m + 15 = 0

m(m – 3) – 5(m -3) = 0

(m – 3) (m – 5) = 0

m = 3, 5

y = C1e3x + C2 e5x

Solve:

Or,

(D2 – 6D + 9)y = 0

Ans. Auxiliary equation are m2 – 6m + 9 = 0

(m – 3)2 = 0 m = 3,3

CF = (C1 + C2x) e3x

Note: If roots are in complex form i.e. α = iβ

CF = eax [ Cos βx + Sin βx]

Solve: (D2 – 4D + 5)y = 0

Ans. Auxiliary equation are m2 – 4m + 8 = 0

m = -2 ± i

CF = e-2x[ C1 Cos x + C2 Sin x]

Solve. (D2 – 3D- 4)y = 0

Ans. Its auxiliary equation is

m2 – 3m – 4 = 0

m2 – 4m + m – 4 = 0

m(m – 4) + 1(m – 4) = 0

(m – 4)(m + 1) = 0

m = 4, - 1

Solution is y = C1 e4x + C2 e-x

Solve. D3 + D – D – 1)y = 0

Ans. The auxiliary equation is

m3 + m2 – m – 1 = 0

(m – 1)(m2 + 2m + 1) = 0

(m – 1)(m2 + 2m + 1) = 0

(m – 1)(m + 1)2 = 0

m = 1, -1, -1

Hence the solution is

y = C1ex + (C2 + C3x )e-x

Rules to find Particular Integral

Case 1:

If, f(a) = 0 then

If, f’(a) = 0 then

Solve:

Ans. Given, (D2 – 6D + 9)y = 5e3x

Auxiliary equation is m2 – 6m + 9 = 0

(m+3)2 = 0

m = -3, -3

CF = (C1 + C2x) e-3x

Put D = 3

CS = CF + PI

= (C1 + C2x) e-3x +

Case2:

Expand by the binomial theorem in ascending powers of D as far as the result of operation on xn is zero.

Solve. (D2 + 5D + 4)y = (3 – 2x)

Given, (D2 + 5D + 4)y = (3 – 2x)

For CF,

Auxiliary equation are m2 + 5m + 4 = 0

= m(m +4) + 1(m+4) = 0

= (m+1)(m+4) = 0

m = -1, -4

CF = C1e-x + C2e-4x

For PI

CS = CF + PI

= C1 e-x + C2e-4x +1/8 (11 – 4x)

Case 3:

Or,

Solve: (D2 + 4)y = Cos 2x

Ans. Auxiliary equation are m2 + 4 = 0

CF = A Cos 2x + B Sin 2x

PI =

Put , D2 = - a2 = -22 = -4

PI =

PI = x

= x/4 Sin 2x

CS = CF + PR

= A Cos 2x + B Sin 2x + x/4 Sin 2x

Case 4:

Solve. (D2 – 4D + 4)y = x3e2x

Ans. AE= m2 – 4m + 4 = 0

(m – 2)2 = 0 m = 2, 2

CF = (C1 + C2 X) e2x

Complete solution is y = (C1 + C2x) e2x + e2x

Solve (9D2 + 12D + 4)y = 6e-2x/3

Ans. The AE is 9m2 + 12m + 4 = 0

(3m + 2)2 = 0

(3m + 2)(3m + 2) = 0

m = -2/3, -2/3

CF = (C1 + C2 x) e-2x/3

Complete solution y= CF + PI

y = (C1 + C2x) e-2x/3 + x2/3

Solve. (D2 + 3D+2)y = 5

Ans. The AE is m2 + 3m + 2 = 0

m2 + m+ 2m + 2 = 0

m(m+1)+2(m+1) = 0

(m+1)(m+2) = 0

m = -1, -2

CF = C1 e-x + C2e-2x

Complete solution = CF + PI

C1e-x + C2e-2x + 5/2

Solve. (D2 – 6D +9)y = 2x2 – x +3

Ans. The AE is m2 – 6m + 9 = 0

(m – 3)2 = 0

m = 3, 3

CF = (C1 + C2x)e3x

Complete solutio0n is y= CF + PI

y – (C + C2x) e3x + 1/9 (2x2 + 5/3x + 11/3)

Find the PI of (D2 + 5D + 4) y = x2 + 7x + 9

Ans.

Solve

Ans. Given equation in symbolic form is

(D2 – 5D + 6)y = sin 3x

Its Auxiliary equation is m2 – 3m + 6 = 0

(m – 2)(m – 3) = 0

m = 2, 3

CF = C1e2x + C2e3x

Complete solution is y= CF + PI

y = C1e2x + C2e3x + 1/78 (5 cos 3x – sin 3x)

Solve. (D2 – 4)y = Cos2x

Ans. The AE is m2 – 4 = 0

m= ± 2

CF = C1 e2x + C2 e-2x

We know, cos 2x = 2 cos2x – 1

Cos2 x = (1+Cos 2x)/2

PI =

(since D2 -22 = -4)

= - 1/8 – 1/16 Cos 2x

Complete solution is y= CF + PI

y = C1e2x + C2e-2x – 1/8 – 1/16 Cos 2x

Solve. Find the PI of (D2 – 4D + 3)y = ex cos 2x

Ans.

Solve. (D3-7D-6) y=e2x (1+x)

Ans. The auxiliary equation i9s

m3 – 7m – 6 = 0

(m+1)(m2 –m – 6) = 0

(m+1)(m – 3)(m +2)=0

m = -1, 3, -2

CF = C1e-x + C2 e3x + C3 e-2x

Hence complete solution is y= CF + PI

y = C1e2x+ C2e3x+ C3e-2x – 1/12 e2x(17/12 + x)

Bessel equations

The Bessel equation is-

The solution of this equations will be-

The Bessel function is denoted by and defined as-

If we put n = 0 then Bessel function becomes-

Now if n = 1, then-

The graph of these two equations will be-

General solution of Bessel equation-

Example: Prove that-

Sol.

As we know that-

Jn(x) =

Now put n = 1/2 in equation (1), then we get-

J1/2(x) =

Hence proved.

Example: Prove that-

Sol.

Put n = -1/2 in equation (1) of the above question, we get-

J1/2(x) =

Recurrence formulae-

Formula-1:

Proof:

As we know that-

On differentiating with respect to x, we obtain-

Putting r – 1 = s

= nJn – x Jn+1

Formula-2:

Proof:

We have-

Differentiating w.r.t. x, we get-

Formula-3:

Proof:

We know that from formula first and second-

Now adding these two, we get-

Or

Formula-4:

Proof:

We know that-

XJ’n = nJn – x Jn-1

XJ’n = - nJn + xJn+1

On subtracting, we get-

0 = 2n Jn – x Jn+1 – x Jn -1

Formula-5:

Proof:

We know that-

Multiply this by we get-

x-n J’n = nx-n-1Jn – x-n Jn+1

I.e.

x-n J’n – n x-n-1 Jn = - x-n Jn+1

Or

Formula-6:

Proof:

We know that-

XJ’n = - n Jn + x Jn-1

Multiply by we get-

xnJ’n = -n xn-1Jn + xnJn-1

xnJ’n + n xn – 1Jn = xn Jn - 1

Or

Example: Show that-

By using recurrence relation.

Sol.

We know that-

The recurrence formula-

On differentiating, we get-

Now replace n by n -1 and n by n+1 in (1), we have-

2 J’n – 1 = Jn – 2 – Jn or J’n – 1 = ½ Jn – 2 – ½ Jn

2 J’n+1 = Jn – Jn+2 or J’n+1 = ½ Jn – ½ Jn+2

Put the values of and from the above equations in (2), we get-

2J’’n = ½ [ Jn-2 – Jn] – ½ [ Jn – Jn + 2]

4J’’n = Jn – 2 – Jn – Jn + Jn+2

4J’’n = Jn – 2 – 2 Jn + Jn+2

Example: Prove that-

Sol.

We know that- from recurrence formula

On integrating we get-

On taking n = 2 in (1), we get-

Again-

Put the value of from equation (2) and (3), we get-

By equation (1), when n = 1

Key takeaways-

- The Bessel function is denoted by and defined as-

2. General solution of Bessel equation-

Legendre’s equations

The Legendre’s equations is-

Now the solution of the given equation is the series of descending powers of x is-

Here is an arbitrary constant.

If n is a positive integer and

If f(x) is a continuous function having finite number of

Oscillations in the interval (0, a), then We can write

The above solution is

So that-

Here is called the Legendre’s function of first kind.

Note- Legendre’s equations of second kind is and can be defined as-

The general solution of Legendre’s equation is-

Here A and B are arbitrary constants.

Rodrigue’s formula-

Rodrigue’s formula can be defined as-

Legendre Polynomials-

We know that by Rodrigue formula-

If n = 0, then it becomes-

If n = 1,

If n = 2,

Now putting n =3, 4, 5……..n we get-

…………………………………..

Where N = n/2 if n is even and N = 1/2 (n-1) if n is odd.

Example: Express in terms of Legendre polynomials.

Sol.

By equating the coefficients of like powers of x, we get-

Put these values in equation (1), we get-

Example: Let be the Legendre’s polynomial of degree n, then show that for every function f(x) for which the n’th derivative is continuous-

Sol.

We know that-

On integrating by parts, we get-

Now integrate (n – 2) times by parts, we get-

Recurrence formulae for -

Formula-1:

NPn = (2n – 1)x Pn-1 – (n -1)Pn-2

Fromula-2:

XP’n – P’n – 1 = nPn

Formula-3:

P’n – xP’n-1 = nPn-1

Formula-4:

P’n+1 – P’n – 1 = (2n+1)Pn

Formula-5:

(x2-1)P’n = n[xPn – Pn-1]

Formula-6:

(x2 – 1)P’n =(n+1)(Pn+1 – xPn)

Generating function for

Prove that is the coefficient of in the expansion of in ascending powers of z.

Proof:

(1 – 2xz+z2)-1/2 = [1 – z(2x – z)]-1//2

Now coefficient of in

Coefficient of in

Coefficient of in

And so on.

Coefficient of in the expansion of equation (1)-

The coefficients of etc. in (1) are

Therefore-

(1 – 2xz + z2)-1/2 = P0(x) + zP1(x) + z2P2(x) + z3P3(x) + . . . + znPn(x) + …

Example: Show that-

P2n(0) = (-1)n

Sol.

We know that

Equating the coefficients of both sides, we have-

Orthogonality of Legendre polynomials-

Proof: is a solution of

…………………. (1)

And

is a solution of-

……………. (2)

Now multiply (1) by z and (2) by y and subtracting, we have-

Now integrate from -1 to +1, we get-

= 0

Example: Prove that-

By using Rodrigue formula for Legendre function.

On integrating by parts, we get-

Now integrating m – 2 times, we get-

Key takeaways-

- The Legendre’s equations is-

2. The general solution of Legendre’s equation is-

Here A and B are arbitrary constants.

3. Rodrigue’s formula can be defined as-

4. Orthogonality of Legendre polynomials-

References:

- E. Kreyszig, “Advanced Engineering Mathematics”, John Wiley & Sons, 2006.
- P. G. Hoel, S. C. Port And C. J. Stone, “Introduction To Probability Theory”, Universal Book Stall, 2003.
- S. Ross, “A First Course in Probability”, Pearson Education India, 2002.
- W. Feller, “An Introduction To Probability Theory and Its Applications”, Vol. 1, Wiley, 1968.
- N.P. Bali and M. Goyal, “A Text Book of Engineering Mathematics”, Laxmi Publications, 2010.
- B.S. Grewal, “Higher Engineering Mathematics”, Khanna Publishers, 2000.