Undetermined coefficients problem

[Gypsumfantastic]

Homework Statement

Solve the following initial value problem

y'' - 5y' +6y = x*exp(2x), y(0) = y'(0) = 0

Homework Equations

The Attempt at a Solution

Ive found the complimentary solution to be r = 3, 2,

Yg = C1*exp(3x) + C2*exp(2x) + Yp

But to find Yp is giving me the problems, using the method of undetermined coefficients, I have tried to so far

1. (Ax^3 + Bx^2 + Cx + D)*exp(2x)
2. (Ax^2 + Bx + C) * exp(2x)
3. (Ax + B)* exp(2x)

but with no luck can anyone help me with what my guess to solve it

 

[AlephZero]

Taking your 3rd attempt, you seem to be using a trial solution of

Yg = C1 exp(3x) + C2 exp(2x) + A x exp(2x) + B exp (2x).

The terms with coefficients C2 and B are the same, so cross out one of them and you should be able to find the other coefficients.

 

[Gypsumfantastic]

Cheers thanks

 

[HallsofIvy]

Homework Statement

Solve the following initial value problem

y'' - 5y' +6y = x*exp(2x), y(0) = y'(0) = 0

Homework Equations

The Attempt at a Solution

Ive found the complimentary solution to be r = 3, 2,

Yg = C1*exp(3x) + C2*exp(2x) + Yp

But to find Yp is giving me the problems, using the method of undetermined coefficients, I have tried to so far

1. (Ax^3 + Bx^2 + Cx + D)*exp(2x)
2. (Ax^2 + Bx + C) * exp(2x)
3. (Ax + B)* exp(2x)

but with no luck can anyone help me with what my guess to solve it

Those are the trial functions? Normally, for a right hand side x exp(2x) you would try y(x)= (Ax+ B) exp(2x) but since exp(2x) is already a solution to the homogeneous equation you need to multiply by x: try y(x)= (Ax^2+ Bx)exp(2x). You don't need the C in (2) but (Ax^2+ Bx+ C)exp(2x) should work: you should get C= 0 using that. Since you don't show HOW you have tried, I can't comment more.