UNIT2
Linear differential equations
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 = , D = ∅(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 examples to understand the concept,
Example1: Solve (4D² +4D 3)y = 
Solution: Auxiliary equation is 4m² +4m – 3 = 0 We get, (2m+3)(2m – 1) = 0 m = , complementary function: CF is A+ B now we will find particular integral, P.I. = f(x) = . = . = . = . = . General solution is y = CF + PI = A+ B . 
Differential operators
D stands for operation of differential i.e.
stands for the operator of integration. stands for operation of integration twice. Thus, 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 be the CF of Putting the value of in equation (1) we get 
It is called auxiliary equation.
Step 3: Roots Real and Different
If are the roots the CF is If are the roots then 
Step 4 Roots Real and Equal
If both the roots are then CF is If roots are 
Example: Solve 
Ans. Given, Here Auxiliary equation is 
Example: Solve 
Or, Ans. Auxiliary equation are Note: If roots are in complex form i.e. 
Example : Solve 
Ans. Auxiliary equation are 
Rules to find Particular Integral
Case 1: If, If, 
Example:Solve  
Ans. Given, Auxiliary equation is  
Case2: Expand by the
 
Case 3: Or,  
Case 4: Example: Solve Ans. AE= Complete solution is  
Example: Solve
Ans. The AE is Complete solution y= CF + PI  
Example: Solve Ans. The AE is Complete solution = CF + PI  
Example: Solve Ans. The AE is Complete solutio0n is y= CF + PI  
Example: Find the PI of Ans.  
Example: solve Ans. Given equation in symbolic form is Its Auxiliary equation is Complete solution is y= CF + PI
 
Example: Solve Ans. The AE is We know, Complete solution is y= CF + PI  
Example: Find the PI of(D24D+3)y=ex cos2x
Ans.  
Example. Solve(D37D6) y=e2x (1+x)
Ans. The auxiliary equation i9s Hence complete solution is y= CF + PI 
Key takeaways
Where a,b,c are the constants. 2. General solution = C.F. +P.I. 3. 4. 5. Roots Real and Equal 6. Roots Real and Different 7. If roots are in complex form i.e. 
A general method of finding the PI of any function
Or Or 
It is the LDE.
The solution will be 
Example: Solve 
Auxiliary equation Complementary function
Complete Solution is

Example: Solve Solution: Auxiliary equation C.F is
[]
The Complete Solution is 
Example Solve Solution: The Auxiliary equation is The C.F is P.I
The Complete Solution is 
Example Solve Solution: The auxiliary equation is The C.F is [Put ]
The Complete Solution is 
Example Solve Solution: The auxiliary equation is The C.F is But The Complete Solution is 
Example Solve Solution: The auxiliary equation is The C.F is
[By parts] The Complete Solution is 
Example Solve Solution: The auxiliary equation is The C.F is Now, And The Complete Solution 
Method of variation of parameters
Consider a second order LDE with constant coefficient given by Then let the complimentary function is given by Then the particular integral is Where u and v are unknown and to be calculated using the formula u= 
Example1: Solve the following DE by using variation of parameters
Sol. We can write the given equation in symbolic form as To find CF It’s A.E. is So that CF is To find PI Here Now Thus PI = = = = = So that the complete solution is 
Example2: Solve the following by using the method of variation of parameters.
Sol. This can be written as C.F. Auxiliary equation is So that the C.F. will be P.I. Here Now Thus PI = = = So that the complete solution is 
An equation of the form
Here X is the function of x, is called Cauchy’s homogeneous linear equation.
Example1: Solve 
Sol. As it is a Cauchy’s homogeneous linear equation. Put Then the equation becomes [D(D1)D+1]y = t or Auxiliary equation So that C.F.= Hence the solution is , we get 
Example2: Solve 
Sol. On putting so that, and The given equation becomes Or it can be written as So that the auxiliary equation is C.F. = Particular integral Where It’s a Leibnitz’s linear equation having I.F.= Its solution will be P.I. = = So that the complete solution is
An equation of the form Is called Legendre’s linear equation. 
Example3: Solve 
Sol. As we see that this is a 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
Cauchy’s homogeneous linear equation 
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 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
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 (n1) if n is odd. 
Legendre Polynomials
We know that by Rodrigue formula
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
We know that On integrating by parts, we get
Now integrate (n – 2) times by parts, we get

Recurrence formulae for 
Formula1: Fromula2: Formula3: Formula4: Formula5: Formula6: 
Generating function for 
Prove that is the coefficient of in the expansion of in ascending powers of z. 
Proof: 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

Example: Show that
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 
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
3. 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.
 T. Veerarajan, “Engineering Mathematics”, Tata McgrawHill, New Delhi, 2010
 Higher engineering mathematics, HK Dass
 BV ramana, higher engineering mathematics