# Find a general solution to $y”+3y’+2y=12x^2$

ordinary differential equations

Find a general solution to $$y''+3y'+2y=12x^2$$

My working out:
Find the roots
$$y^2+3y+2 = 0 \implies (y+1)(y+2)=0 \\ =c_1e^{-2x}+c_2e^{-x}$$
Using the method of undetermined coefficients, we have

$$r(x) = 12x^2 \implies y_p = K_2x^2 \\ y_p' = 2K_1x \\ y_p'' = 2K_0$$

Equating the coefficients
$$K_2x^2+6K_1x+4K_0 = 12x^2$$

How do I proceed from here?

I know that if $$K_1 = K_0 = 0$$ I have $$K_2x^2=12x^2$$ however I don't know how to get the final answer from here which is $$y_h+y_p = c_1e^{-2x}+c_2e^{-x}+6x^2-18x+21$$?

Using the additional comment, here's what I have tried:

$$y_p = K_2x^2 + K_1x + K_0 \\ y'_p = 2K_2x+K_1 \\ y'' = 2K_2$$

$$\implies 2K_2+3(2K_2x+K_1) + 2(K_2x^2+K_1x+K_0) = 12x^2$$
We have that $$12+3K_1+2K_0 = 0 \\ 36x+2K_1=0 \\ \implies K_2 = 6, K_1 = -18, K_0 = 21$$

You are 10 seconds away from the solution. Your mistake is assuming the wrong particular solution.

In general for particular solution,

$$y_p=\sum_{i=0}^n K_nx^n,$$ where $$n$$ is the highest order of the polynomial.

$$y_p=K_2x^2+K_1x+K_0$$

Now, follow the above $$y_p$$,

$$y_p^\prime=2K_2x+K_1\\ y_p^{\prime\prime}=2K_2$$

Obviously $$K_2=6$$. We also have $$2K_0+3K_1+2K_2=0$$. There is one more equation, binding $$K_2$$ to $$K_1$$.

I hope this helps