Problem Statement
So here is the problem.
Solve $$ xy'' + 2y' = 4x^3 $$
Note there is no boundary conditions provided here though.
The textbook provided a solution using reduction of order (which is attached below), but I tried to solve it with Laplace Transform, which gives a slightly different form. Basically, the two solutions are,
From reduction of order, $y=\frac{1}{5}x^4 -\frac{C_1}{x} + C_2$
From Laplace Transform, $y=\frac{1}{5}x^4 + C_1\delta(x)+C_2$
So you see, the second terms are different from two solutions. One gives a inverse proportion as $\frac{1}{x}$, while the other is a delta function.
They look similar (i can sort of perceive) but are both solutions correct? If so, how could I consolidate the two solutions? I am hoping to get more understanding into the two methods. Thanks!
The detailed solution steps are attached below.
Attached Detailed Solutions
Reduction of order
Let $u=y'$, so that, $$ x\frac{du}{dx} + 2u = 4x^3 $$ Multiplying by x, $$ x^2\frac{du}{dx} + 2xu = 4x^4\\ \frac{d}{dx}(x^2u) = 4x^4 $$ This can be integrated to give $$ u = \frac{4}{5}x^3 + \frac{C_1}{x^2} $$ from which, $$ y = \frac{1}{5}x^4 - \frac{C_1}{x} + C_2 $$ for $x\neq0$
Laplace Transform
First apply Laplace transform to all terms, $$ \mathcal{L}(xy'') = -2sY - s^2Y' + y(0)\\ \mathcal{L}(y') = sY - y(0)\\ \mathcal{L}(x^3) = \frac{3!}{s^4} $$ Putting everything into the ODE and organize, we have, $$ Y' = -24s^{-6} - y(0)s^{-2}\\ Y = \frac{24}{5}s^{-5} + y(0)s^{-1} + C $$ Apply inverse Laplace transform, we can get, $$ y = \frac{1}{5}x^4 + y(0) + C\delta(t) $$ Rename constants to a comparable form, we have finally $$ y = \frac{1}{5}x^4 + C_1\delta(t) + C_2 $$ For this case, $x>=0$.
HINT
This answer has nothing to do with your question, but provides another way to solve the problem.
\begin{align*} xy^{\prime\prime} + 2y^{\prime} = 4x^{3} \Longleftrightarrow x^{2}y^{\prime\prime} + 2xy^{\prime} = 4x^{4} \Longleftrightarrow (x^{2}y^{\prime})^{\prime} = 4x^{4} \Longleftrightarrow x^{2}y^{\prime} = \frac{4x^{5}}{5} + k \end{align*}