 # Second degree linear derivate with exponential function, which general solution? Suppose James Dale 2022-01-17 Answered
Second degree linear derivate with exponential function, which general solution?
Suppose $y-4y=x{e}^{2x}$
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For any polynomial P, constant k and function u, $P\left(D\right){e}^{kx}u=\mathrm{exp}\left(kx\right)P\left(D+k\right)u$ (where D stands for derivative). In your case . Now you want . Writing $v={u}^{\prime }$, we have the first-order linear equation in $v:{v}^{\prime }+4v=x$, which has a solution $v=\frac{x}{4}-\frac{1}{16}$. An antiderivative of this is $u=\frac{{x}^{2}}{8}-\frac{x}{16}$, corresponding to the particular solution $y={e}^{2x}\left(\frac{{x}^{2}}{8}-\frac{x}{16}\right)$ of your original equation.

We have step-by-step solutions for your answer! Medicim6
One way to get this mechanically is to use the method of Laplace Transforms.
Of course, I havent

We have step-by-step solutions for your answer! alenahelenash

You want to solve an equation $Ly={x}^{n}{e}^{\lambda x}$, where L is a constant linear differential operator - in your case $Ly={y}^{″}-4y$. Let us solve $Ly={e}^{\left(\lambda +ϵ\right)x}$ instead: if we substitute $y=c\left(ϵ\right){e}^{\left(\lambda +ϵ\right)x}$, we get $c\left(ϵ\right)p\left(\lambda +ϵ\right)=1$ (where p is the characteristic polynomial of L, in your case $p\left(t\right)={t}^{2}-4\right),i.e.c\left(ϵ\right)=1/p\left(\lambda +ϵ\right)$. We now expand
$L\frac{{e}^{\left(\lambda +ϵ\right)x}}{p\left(\lambda +ϵ\right)}={e}^{\left(\lambda +ϵ\right)x}$
to a power series in $ϵ$, look at the term at ${ϵ}^{n}$, and get a solution of $Ly={x}^{n}{e}^{\lambda x}/n!$.
In your case . We thus get
$L\left(\left({ϵ}^{-1}{4}^{-1}+\left({4}^{-1}x-{4}^{-2}\right)+ϵ\left({4}^{-1}{x}^{2}/2-{4}^{-2}x+{4}^{-3}\right)+\cdots \right){e}^{2x}\right)=\left(1+ϵx+{ϵ}^{2}{x}^{2}/2+\cdots \right){e}^{2x}$
Looking at the term at $ϵ$, we get
$L\left(\left({4}^{-1}{x}^{2}/2-{4}^{-2}x+{4}^{-3}\right){e}^{2x}\right)=x{e}^{2x}$,
i.e. $y=\left({4}^{-1}{x}^{2}/2-{4}^{-2}x+{4}^{-3}\right){e}^{2x}$ is a solution of your equation (we can drop ${4}^{-3}{e}^{2x}$ from the solution, as it solves the homogeneous equation).

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