Solving a Differential Equation: Finding the Particular Solution

[mansfin]

I have to find the particular solution to the differential equation:

(-21/4)y''+2y'+y=4xe^(3x)

First, I chose my trial function to be yp=(Ax+B)*e^(3x). Is this correct?
so yp'=3(Ax+B)*e^(3x)
yp''=9(Ax+B)*e^(3x)

So I plug these into the differential equation and I get:
(-189/4)Axe^(3x)-(189/4)Be^(3x)+6Axe^(3x)+6Be^(3x)+Axe^(3x)+Be^(3x)=4xe^(3x)

I group like terms and I get A=-16/161 and B=0
So yp=(-16/161)Axe^(3x)

This is not correct.
Can someone please tell me where I'm going wrong? Thanks!

 

[Mark44]

I have to find the particular solution to the differential equation:

(-21/4)y''+2y'+y=4xe^(3x)

First, I chose my trial function to be yp=(Ax+B)*e^(3x). Is this correct?
so yp'=3(Ax+B)*e^(3x)
yp''=9(Ax+B)*e^(3x)

So I plug these into the differential equation and I get:
(-189/4)Axe^(3x)-(189/4)Be^(3x)+6Axe^(3x)+6Be^(3x)+Axe^(3x)+Be^(3x)=4xe^(3x)

I group like terms and I get A=-16/161 and B=0
So yp=(-16/161)Axe^(3x)

This is not correct.
Can someone please tell me where I'm going wrong? Thanks!

The particular solution you choose is affected by the solutions to the homogeneous equation. For your problem, I'm going to guess that e^3x is not a solution to the homogeneous problem, but I don't know that for a fact.

In any case, and assuming that y = e^3x is not a solution to the homogeneous problem, your choice for a particular solution is good, but you made a mistake in both of your derivatives. differentiation. =(Ax+B)*e^3x is a product, a fact that you seem to have completely overlooked.

 

[mansfin]

Wow. Implied multiplication on my calculator. Next time I will just work that out by hand. Thanks for your help!