How to Solve a Second Order Nonhomogeneous Differential Equation?

[Math10]

Homework Statement

Find a particular solution of y"-5y'+6y=-e^(x)[(4+6x-x^2)cosx-(2-4x+3x^2)sinx].

Homework Equations

None.

The Attempt at a Solution

yp=e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
y'p=e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
y"p=e^(x)[(Ax^2+Bx+C)(-cosx+sinx)+(2Ax+B)(-sinx-cosx)+(2Ax+B)(-sinx-cosx)+2A(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
subbing:
y"p-5y'p+6yp=4Ax^2e^xcosx+4Bxe^xcosx+4Ce^xcosx+2Ax^2e^xsinx+2Bxe^xsinx+2Ce^xsinx+2Axe^xsinx+Be^xsinx-10Axe^xcosx-5Be^xcosx+2Ae^xcosx-2Ae^xsinx=-e^x[(4+6x-x^2)cosx-(2-4x+3x^2)sinx]
So 4Ax^2e^xcosx=x^2e^xcosx
and therefore A=1/4.
But the answer to this problem is yp=e^(x)(x^2*cosx+2sinx). I don't think the coefficients I got are right. So what's the right initial guess for this problem?

 

[vela]

How did you come up with your guess?

 

[Math10]

Just by making my best guess.

 

[vela]

What made you think it was the best as opposed to some other form? I'm asking because writing down the form of particular solution is a cookbook process, yet you managed to come up with something different.

 

[Math10]

Since there's -e^x in the very front, so I got e^x as well in forming my guess yp. And then there's (4+6x-x^2) in front of cosx and same thing in front of sinx. So I got yp=e^(x)[(Ax^2+Bx+C)(cosx-sinx)].

 

[vela]

The sine doesn't have the same factor multiplying it as cosine in the forcing function, or are you saying the cosine and the sine should both have a quadratic multiplying it? If the latter, why would you expect it to be the exact same quadratic?

 

[Math10]

So this is a forcing function?

 

[vela]

The forcing function is the stuff that appears on the righthand side of the differential equation.

 

[Math10]

So what's the way to find the guess?

 

[Ray Vickson]

Homework Statement

Find a particular solution of y"-5y'+6y=-e^(x)[(4+6x-x^2)cosx-(2-4x+3x^2)sinx].

Homework Equations

None.

The Attempt at a Solution

yp=e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
y'p=e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
y"p=e^(x)[(Ax^2+Bx+C)(-cosx+sinx)+(2Ax+B)(-sinx-cosx)+(2Ax+B)(-sinx-cosx)+2A(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(-sinx-cosx)+(2Ax+B)(cosx-sinx)]+e^(x)[(Ax^2+Bx+C)(cosx-sinx)]
subbing:
y"p-5y'p+6yp=4Ax^2e^xcosx+4Bxe^xcosx+4Ce^xcosx+2Ax^2e^xsinx+2Bxe^xsinx+2Ce^xsinx+2Axe^xsinx+Be^xsinx-10Axe^xcosx-5Be^xcosx+2Ae^xcosx-2Ae^xsinx=-e^x[(4+6x-x^2)cosx-(2-4x+3x^2)sinx]
So 4Ax^2e^xcosx=x^2e^xcosx
and therefore A=1/4.
But the answer to this problem is yp=e^(x)(x^2*cosx+2sinx). I don't think the coefficients I got are right. So what's the right initial guess for this problem?

Have you checked whether the supposed "answer" actually solves the DE?