Solve :(adsbygoogle = window.adsbygoogle || []).push({});

[tex]y'' + y' - 12y = 4x^2[/tex]

The complementary equation I get is [tex] y1 = C1 e^3x + C2 e^-4x [/tex]

But how to solve for the trial solution?

I do it in this way:

[tex] f(x) = 4x^2 [/tex]

[tex] y2 = D (Ax^2 + Bx + C )[/tex]......

What I want to know is whether my y2 is correct.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Differential Equations

**Physics Forums | Science Articles, Homework Help, Discussion**