Find a solution to $xy'+y = x^k$ using the Laplace transform.

Question

The part that is giving me trouble is:

$$\mathcal{L}[xy']$$

I have never done this before so I ploughed through.

$$\mathcal{L}[xy'] = \int^{\infty}_0ty'(t)e^{-pt}dt = -\frac{d}{dp}\int^{\infty}_0y'(t)e^{-pt}dt = -\frac{d}{dp}\mathcal{L}[y'] = \frac{d}{dp}[y(0) - \mathcal{L}[y]]$$

so in the grand scheme of things $$\frac{d}{dp}[y(0) - \mathcal{L}[y]] + \mathcal{L}[y] = \frac{k!}{p^{k+1}}$$

How can I proceed?

Could I have done something differently to get something nicer?

$$\color{red}{\text{The restriction that I use a Laplace transform is necessary. }}$$

Answer 1

After a few trial and errors I came up with this;

If we make the substitution $x = e^z$

then $$xy'=x\frac{dy}{dx} = e^z\frac{dy}{dz}(\frac{dy}{dx})^{-1}=\frac{dy}{dz}$$

so the equation in question becomes:

$$y' + y = e^{kz}$$

on taking the Laplace transform of both sides

$$p\mathcal{L}[y]-y_0 + \mathcal{L}[y] = \frac{1}{p-k}$$

$$\implies \mathcal{L}[y] = \frac{1+y_0(p-k)}{(p-k)(p+1)}$$

$$\implies \mathcal{L}[y] = \frac{1}{(p-k)(p+1)}+\frac{y_0}{(p+1)}$$

$$\implies y = y_0e^{-z} + \frac{e^{-z}+e^{kz}}{1+k}$$

On making our substitutions

$$y = \frac{y_0}{x}+\frac{\frac{1}{x}+x^k}{k+1}$$

It would be great if someone could refute/confirm the validity of my solution.

It does satisfy the differential equation.

Answer 2

Now you have a differential equation for $\mathcal L[y]$. The whole purpose of the Laplace Transform is to change differential equations into algebraic equations. Here all you've done is trade one differential equation for another, so it doesn't seem like the Laplace transform is the ideal solution method. Is there a particular reason you are using the Laplace transform, rather than, say, an integrating factor? The equation already has the form $$ \frac{d}{dx}[xy(x)] = x^k $$ so just integrating, we see $$ xy(x) = \frac{x^{k+1}}{k+1} + C, $$ so $$ y(x) = \frac{x^k}{k+1} + \frac{C}{x}. $$ Then you can solve for $C$ using a condition of the form $y(x_0) = y_0$ for some $x_0 \neq 0$.