<s>[INST] solve: $ \frac {d^2y}{dx^2} + 5\frac{dy}{dx} = 15x^2$ $$ \frac {d^2y}{dx^2} + 5\frac{dy}{dx} = 15x^2 $$
It's solution is
$$y = y_{h} + y_{p}$$
where $y_{h} $ is the solution for homogenous equation and $y_{p} $ is the particular solution
For homogenous solution:
It's auxiliary equation is
$$m^2 + 5m = 0$$
on solving for m,
$$m = 0, -5$$
so, $$y_{h} = C_{1} + C_{2} e^{-5x}$$
here
$$R = 15x^2$$
so it's particular solution should be :
$$y_{p} = K_{2}x^2 + k_{1}x + k_{0}$$
on differentiating with respect to x
$$y\prime _{p}= 2xk_{2} + k_{1}$$
again
$$y\prime\prime_{p} = 2k_{2}$$
using all above values in the initial equation gives
$$2k_{2} + 10xk_{2} + 10k_{1} = 15x^2$$
But using this method I am unable to proceed for the value of all unknown especially $k_{0}$
The given answer is:
$$y = C_{1} + C_{2}e^{-5x} + x^3 + \frac{3x^2}{5} - \frac{6}{25}x$$
 [/INST] HINT
Notice the proposed equation is equivalent to a first order ODE:
\begin{align*}
y'' + 5y' = 15x^{2} & \Longleftrightarrow y' + 5y = 5x^{3} + c_{0}\\\\
& \Longleftrightarrow \exp(5x)y' + 5\exp(5x)y = 5\exp(5x)x^{3} + c_{0}\exp(5x)\\\\
& \Longleftrightarrow [\exp(5x)y]' = 5\exp(5x)x^{3} + c_{0}\exp(5x)\\\\
& \Longleftrightarrow \exp(5x)y = 5\int\exp(5x)x^{3}\mathrm{d}x + \frac{c_{0}}{5}\exp(5x) + c_{1}\\\\
& \Longleftrightarrow y = 5\exp(-5x)\int\exp(5x)x^{3}\mathrm{d}x + \frac{c_{0}}{5} + c_{1}\exp(-5x)
\end{align*}
Can you take it from here?
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