How to Solve for the Trial Solution in Differential Equations?

[frozen7]

Solve :

y'' + y' - 12y = 4x^2
The complementary equation I get is y1 = C1 e^3x + C2 e^-4x

But how to solve for the trial solution?
I do it in this way:

f(x) = 4x^2
y2 = D (Ax^2 + Bx + C )...
What I want to know is whether my y2 is correct.

 

[Tide]

Are you familiar with the method of variation of parameters?

 

[TD]

You can certainly use undetermined coefficients since your RHS is just a polynomial. Since its degree is 2, your suggestion for a solution should be a general second degree polynomial as well.

Your y2 is fine, but it can be simplified a bit, the D isn't necessary. If you would work it out, you'd get ADx²+BDx+CD where 'AD', 'BD' and 'CD' are again just 3 constants so just using A, B and C is fine - then you have the most general second degree polynomial.

Find its first and second derivative, plug it into your DE and identify coefficients to solve for A, B and C

 

[mathmike]

i think you should try variation of parameters as tide suggested

 

[frozen7]

What is variation of parameters??

 

[TD]

What is variation of parameters??

You can google it, the method is described well in http://www.math.utah.edu/~gustafso/2250variation-of-parameters.pdf" .
This is certainly a method worth learning since it applies more generally than the method of undetermined coefficients (which only works for a limited number of RHS functions)

Just as a note: your method (undetermined coefficients) will work here as well!

 

[frozen7]

Thanks a lot.

 

[TD]

You're welcome, don't hesitate to ask for help if you're stuck