- #1

Math10

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- 0

## 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?