Problem source

Let $P,Q$ and $R$ be functions of $x$. Prove that, for any function
$y$ of $x$, the function 
\[
Py''+Qy'+Ry
\]
can be written in the form
$\dfrac{\mathrm{d}}{\mathrm{d}x}(py'+qy),$ where $p$ and $q$ are
functions of $x$, if and only if $P''-Q'+R=0.$ 
Solve the differential equation 
\[
(x-x^{4})y''+(1-7x^{3})y'-9x^{2}y=(x^{3}+3x^2)\mathrm{e}^{x},
\]
given that when $x=2,y=2\mathrm{e}^{2}$ and $y'=0.$

Solution source

Suppose $Py'' + Qy' + Ry = \frac{\d}{\d x}(p y' + qy)$, then

\begin{align*}
Py'' + Qy' + Ry &= \frac{\d}{\d x}(p y' + qy) \\
&= py'' + p'y' + qy' + q' y \\
&= py'' + (p'+q)y' + q' y 
\end{align*}

Therefore $P = p, Q = p'+q, R = q'$, Therefore $q = Q-P'$ and $R = Q'-P''$ or $P'' -Q'+R = 0$.

$(\Rightarrow)$ Suppose it can be written in that form, then the logic we have applied shows that equation is true.
$(\Leftarrow)$ Suppose $P''-Q'+R = 0$, then letting $p = P, q = Q-P'$ we find functions of the form which will be expressed correctly.


Notice that if $P = x-x^4, Q = (1-7x^3), R = -9x^2$ then:

\begin{align*}
P'' - Q' + R &= -12x^2+21x^2-9x^2 \\
&= 0
\end{align*}

Therefore we can write our second order ODE as:

\begin{align*}
&& (x^{3}+3x)\mathrm{e}^{x} &= \frac{\d}{\d x} \left ( (x-x^4) y'  +(1-7x^3-(1-4x^3))y \right) \\
&&&= \frac{\d}{\d x} \left ( (x-x^4) y'  -3x^3y \right)
\end{align*}

Suppose $z = (x-x^4)y' - 3x^2y$, then $z = (2-2^4) \cdot 0 - 3 \cdot 2^2 \cdot 2e^2 = -24e^2$ when $x = 2$. and we have:

\begin{align*}
&& \frac{\d z}{\d x} &= (x^3+3x^2)e^x \\
\Rightarrow && z &= \int (x^3+3x^2)e^x \d x \\
&&&= x^3 e^x+c \\
\Rightarrow && -48e^2 &= e^2(8) + c \\
\Rightarrow && c &= -56e^2 \\
\Rightarrow && z &= e^x(x^3)-56e^2 \\
\end{align*}

So our differential equation is:

\begin{align*}
&& (x-x^4)y'-3x^3 y &= x^3e^x -5 6e^2  \\
\Rightarrow && (1-x^3)y' -3x^2y &= x^2 e^x - \frac{6e^2}{x} \\
\Rightarrow && \frac{\d }{\d x} \left ( (1-x^3)y \right) &= x^2e^x - \frac{56e^2}{x} \\
\Rightarrow && (1-x^3)y &= (x^2-2x+2)e^x - 56e^2 \ln x + k \\
\underbrace{\Rightarrow}_{x=2} &&   (1-2^3)2e^2 &= (2^2-2\cdot2 + 2)e^2 -56e^2 \ln 2 + k \\
\Rightarrow && k &= -16e^2+56 \ln 2 \cdot e^2 \\
\Rightarrow && y &= \frac{(x^2-2x+2)e^x - 56e^2 \ln x -16e^2+56 \ln 2 \cdot e^2}{(1-x^3)}
\end{align*}