how to solve this ODE question??

Question

this is the question....
y"=3y'+2y+sinh x

Please help me~...! :-)

Answer 1

y" - 3y' - 2y = sinh x
this non-homogeneous DE may be written as:
y" - 3y' - 2y = (1/2) [exp(x) - exp(-x) ] (1)

The first term of this DE is solvable as homogeneous with a complimentary solution, and second term as a particular solution, and then combined.

(M) Homogeneous with a complimentary solution
y" - 3y' - 2y =0
D^2 – 3D – 2=0 or roots
[D – (½){3+sqrt(17)}] [(D - (½){3-sqrt(17)}] =0

y (c) = c1 exp[(½){3+sqrt(17)}] x + c2 exp[(½){3-sqrt(17)}] x ...(2)

(N) particular solution (for right hand term)

y (p) has to be guessed. It looks like that

y (particular) = A exp(x) + B exp(-x) (3)

will work: find derivatives and putting in (1)

y" = A exp(x) + B exp(-x) & - 3y' = -3 [A exp(x) - B exp(-x)] &
- 2y = -2 [A exp(x) - B exp(-x)]

(1) gives - 4 A [exp(x)] + 6 B [exp(-x)] = (1/2) [exp(x) - exp(-x)]

comparing coefficients of exp(x) and exp(-x)
we get - 4 A= 1/2 6 B = - 1/2
put A= -1/8 B = - 1/12 in (3)
y (p) = -1/8 exp(x) - 1/12 exp(-x)

The general solution to (1) is

y = c1 exp[(½){3+sqrt(17)}] x + c2 exp[(½){3-sqrt(17)}] x
-1/8 exp(x) - 1/12 exp(-x)

Using some boundary conditions, c1 and c2 can be found.

Answer 2

1)
Solve y"=3y'+2y first, solution is in the form y = Aexp(λx).
Solve quadratic equation for λ=λ1, λ2

2)
Find any solution for y"=3y'+2y+sinh x, solution is in the form y = C sinh(x) + D cosh(x).

3)
Combine 1) and 2)