Question

y''-2y'-3=e^4x

Answer 1

Do you mean \(y''-2y'-3y=e^{4x}\), or is it as you typed it? I'm going to assume no typos.

First rearrange:
\[y''-2y'=e^{4x}+3\]
Homogeneous solution:
\[y''-2y'=0\]
gives the characteristic polynomial,
\[r^2-2r=0\]
which has roots \(r=0,~r=2\), and so the general homogeneous solution must be \(y=C_1+C_2e^{2x}\).

For the nonhomogeneous part, we can use the method of undetermined coefficients and use a guess solution of \(y=Ae^{4x}+Bx^2+Cx+D\), then
\[\begin{align*}
y&=Ae^{4x}+Bx^2+Cx+D\\
y'&=4Ae^{4x}+2Bx+C\\
y''&=16Ae^{4x}+2B
\end{align*}\]
Substitute into the equation:
\[\begin{align*}
y''-2y'&=e^{4x}+3\\
16Ae^{4x}+2B-2\left(4Ae^{4x}+2Bx+C\right)&=e^{4x}+3\\
8Ae^{4x}+2B-4Bx-2C&=e^{4x}+3
\end{align*}\]
which gives \(A=\dfrac{1}{8}\), \(B=0\), and \(C=-\dfrac{3}{2}\).

The general solution is then
\[y=\frac{1}{8}e^{4x}-\dfrac{3}{2}x+C_1+C_2e^{2x}\]
(The \(D\) from the non-homogeneous guess is absorbed into the constant \(C_1\).)