$2x^2y''+3xy'-\left(x+1\right)y=0$

Question

I am trying to solve $2x^2y''+3xy'-\left(x+1\right)y=0$ for a while but I can't make a progress.

For a solution near $x=0$ I apply method of series expansion but it doesn't show any closed pattern to related $A_n$ to $A_0$ ; and by more 'direct' method (for $x \ne 0$) it can be reduced to $(x^{3/2} y')'=\dfrac{x+1}{2 \sqrt x} y$ but I don't know how to go further: On one hand there is $y$ is RHS and on the other hand, there is $x^{3/2}$ multiplied by $y'$ in LHS which both ways makes it impossible to proceed.

However, there is a closed form (suggested by wolframalpha - without steps) for the solution : $$y(x) = c_1 \frac1x e^{\sqrt {2x}} (\sqrt 2 - 2 \sqrt x) + c_2 \frac1x e^{-\sqrt {2x}} ( 1+ \sqrt {2x}).$$

Answer 1

Apply the Laplace transformation:

$$\mathcal{L}_x\left[2x^2\frac{\mathrm{d}^2y(x)}{\mathrm{d}x^2}+3x\frac{\mathrm{d}y(x)}{\mathrm{d}x}-y(x)-xy(x)\right](s)=\mathcal{L}_x[0](s)$$

$$\implies 2\frac{\mathrm{d^2}}{\mathrm{ds^2}}\left(\mathcal{L}_x\left[\frac{\mathrm{d}^2y(x)}{\mathrm{d}x^2}\right](s)\right)-3\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathcal{L}_x\left[\frac{\mathrm{d}y(x)}{\mathrm{d}x}\right](s)\right)-\mathcal{L}_x[y(x)](s)+\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)=0$$

$$\implies \color{darkgreen}{2\frac{\mathrm{d^2}}{\mathrm{ds^2}}\left(s^2\mathcal{L}_x\left[y(x)\right](s)-sy(0)-y'(0)\right)-3\frac{\mathrm{d}}{\mathrm{ds}}\left(s\mathcal{L}_x\left[y(x)\right](s)-y(0)\right)-\mathcal{L}_x[y(x)](s)+\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)=0\tag{1}}$$

Now they take:

$$\frac{\mathrm{d}}{\mathrm{ds}}\left(s\mathcal{L}_x\left[y(x)\right](s)-y(0)\right)=s\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)+\mathcal{L}_x[y(x)](s)$$ $$\frac{\mathrm{d^2}}{\mathrm{ds^2}}\left(s^2\mathcal{L}_x\left[y(x)\right](s)-sy(0)-y'(0)\right)=s^2\frac{\mathrm{d^2}}{\mathrm{ds^2}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)+4s\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)+2\mathcal{L}_x[y(x)](s)$$

Substituting in $\color{darkgreen}{(1)}$, we have:

$$\implies 2\left(s^2\frac{\mathrm{d^2}}{\mathrm{ds^2}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)+4s\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)+2\mathcal{L}_x[y(x)](s)\right)-3\left(s\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)+\mathcal{L}_x[y(x)](s)\right)-\mathcal{L}_x[y(x)](s)+\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)=0$$

Simplifying, we have:

$$\implies 2s^2\frac{\mathrm{d^2}}{\mathrm{ds^2}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)+(5s+1)\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)=0$$

Let $\displaystyle\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathcal{L}_x\left[y(x)\right](s)\right)=u(s)$, which gives $\displaystyle\frac{\mathrm{d}^2}{\mathrm{d}s^2}\left(\mathcal{L}_x\left[y(x)\right](s)\right)=\frac{\mathrm{d}(u(s))}{\mathrm{d}s}$, Now solve for $\displaystyle\frac{\mathrm{d}(u(s))}{\mathrm{d}s}$:

$$\implies \frac{\mathrm{d}u(s)}{ds}=\frac{(-5s-1)u(s)}{2s^2}$$

Divide both sides by $\displaystyle \frac{u(s)}{2}$:

$$\implies \frac{2\frac{\mathrm{d}u(s)}{ds}}{u(s)}=\frac{-5s-1}{s^2}$$

Integrate both sides with respect to s:

$$\implies \int\frac{2\frac{du(s)}{ds}}{u(s)}\mathrm{d}s=\int\frac{-5s-1}{s^2}\mathrm{d}s$$ $$\implies 2\log(u(s))=\frac{1}{s}-5\log(s)+c_1$$

Where $c_1$ is an arbitrary constant.

$$\implies u(s)=\frac{c_1e^{0.5/s}}{s^{5/2}}$$

Substitute back $\displaystyle\frac{\mathrm{d}}{\mathrm{d}s}\left(\mathcal{L}_x\left[y(x)\right](s)\right)=u(s)$

$$\implies \frac{\mathrm{d}}{\mathrm{d}s}\left(\mathcal{L}_x\left[y(x)\right](s)\right)=\frac{c_1e^{0.5/s}}{s^{5/2}}$$

Integrate both sides with respect to s:

$$\implies \mathcal{L}_x\left[y(x)\right](s)=\int\frac{c_1e^{0.5/s}}{s^{5/2}}\mathrm{d}s=c_1\left(\frac{2e^{0.5/s}}{\sqrt{s}}+\sqrt{2\pi}\mathrm{erfi}\left(\frac{1}{\sqrt{2}\sqrt{s}}\right)\right)+c_2$$

Where $c_2$ is an arbitrary constant and $\mathrm{erfi}(.)$ is imaginary error function.

Compute:

$$\implies y(x)=\mathcal{L}^{-1}_x\left[\frac{2e^{0.5/s}c_1}{\sqrt{s}}+\sqrt{2\pi}\mathrm{erfi}\left(\frac{1}{\sqrt{2}\sqrt{s}}\right)c_1+c_2\right]$$

$$\implies y(x)=\frac{-2c_1\cosh(\sqrt{2}\sqrt{x})}{\sqrt{\pi}\sqrt{x}}+\sqrt{\frac{2}{\pi}}\frac{c_1\sinh(\sqrt{2}\sqrt{x})}{x}+c_2\delta(x)$$

Hence the solution is:

$$\color{darkgreen}{y(x)=-\frac{2c_1\cosh(\sqrt{2}\sqrt{x})}{\sqrt{x}}+\frac{\sqrt{2}c_1\sinh(\sqrt{2}\sqrt{x})}{x}+c_2\delta(x)}$$

Where $\delta(x)$ is Driac delta function.

Answer 2

$x=0$ is a regular singular point. The indicial equation is $r^2 + r/2 - 1/2 = 0$ which has roots $1/2$ and $-1$, so there are power series solutions of the form $\sum_{k=0}^\infty a_k x^{1/2+k}$ and $\sum_{k=0}^\infty b_k x^{-1+k}$. The recurrences for $a_k$ and $b_k$ are $$ \eqalign{a_{k+1} &= \frac{a_k}{(2k+5)(k+1)} \cr b_{k+1} &= \frac{b_k}{(2k-1)(k+1)} \cr}$$ Thus with $a_0 = 1$ the first series is $$ \eqalign{&x^{1/2} \left(1 + \frac{x}{5 \cdot 1!} + \frac{x^2}{7 \cdot 5 \cdot 2!} + \frac{x^3}{9 \cdot 7 \cdot 5 \cdot 3!} + \ldots\right)\cr&= x^{1/2} \left(1 + \frac{3 \cdot 2 x}{5 \cdot 3!} + \frac{3 (2 x)^2}{7 \cdot 5!} + \frac{3 (2x)^3}{9 \cdot 7!} + \ldots\right)\cr &= \frac{3 \cosh(\sqrt{2x})}{2 x^{1/2}} - \frac{3 \sqrt{2} \sinh(\sqrt{2x})}{4 x}} $$ Similarly, with $b_0 = 1$ the second series turns out to be $$ \frac{\cosh(\sqrt{2x})}{x} - \frac{\sqrt{2} \sinh(\sqrt{2x})}{\sqrt{x}}$$