Unit - 3

Ordinary differential equations of higher orders

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

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

Ordinary differential equations of higher orders

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

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

Ordinary differential equations of higher orders

Unit - 3

Ordinary differential equations of higher orders

Unit - 3

Ordinary differential equations of higher orders

Unit - 3

Ordinary differential equations of higher orders

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)

Unit - 3

Ordinary differential equations of higher orders

Unit - 3

Ordinary differential equations of higher orders

Higher order linear differential equations with constant coefficients

Second order linear homogeneous equations with constant coefficients

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

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)

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

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.