Solving differential equation

Question

I want to solve the following differential equation with initial conditions:

$$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\frac{x \, y(x)}{\sqrt{1-x}}$$ But do not know how to actually solve it. Any suggestion?

Answer 1

One thing you could do is look for a power series solution. This is a linear ODE with an irregular singularity at $x=1$, but if we look for a solution centered at $x=0$, so $|x|<1$, then we should be okay. Note this makes sense to do given where $\frac{x}{\sqrt{1-x}}$ is defined:

Suppose $y = \sum_na_nx^n\Rightarrow y'' = \sum_n n(n-1)a_n x^{n-2}$

Also, $\frac{x}{\sqrt{1-x}} = \sum_{k=0}^\infty {-\frac{1}{2} \choose k} x^{k+1}$, by the binomial series.

Now we can multipy these expressions and solve for the coefficients $a_n$:

$y'' =\displaystyle \sum_n n(n-1)a_n x^{n-2}=\left(\displaystyle\sum_{k=0}^\infty {-\frac{1}{2} \choose k} x^{k+1}\right) \left(\displaystyle\sum_na_nx^n\right)$.

It is messy, but should work out with a bit of computation. Hopefully , it suffices for your needs.

Answer 2

Hint:

Let $u=\sqrt{1-x}$ ,

Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-\dfrac{1}{2\sqrt{1-x}}\dfrac{dy}{du}$

$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-\dfrac{1}{2\sqrt{1-x}}\dfrac{dy}{du}\right)=-\dfrac{1}{2\sqrt{1-x}}\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}=-\dfrac{1}{2\sqrt{1-x}}\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}=-\dfrac{1}{2\sqrt{1-x}}\dfrac{d^2y}{du^2}\left(-\dfrac{1}{2\sqrt{1-x}}\right)+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}=\dfrac{1}{4(1-x)}\dfrac{d^2y}{du^2}+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}$

$\therefore\dfrac{1}{4(1-x)}\dfrac{d^2y}{du^2}+\dfrac{1}{4(1-x)^\frac{3}{2}}\dfrac{dy}{du}=\dfrac{xy}{\sqrt{1-x}}$

$\dfrac{1}{4u^2}\dfrac{d^2y}{du^2}+\dfrac{1}{4u^3}\dfrac{dy}{du}-\dfrac{(1-u^2)y}{u}=0$

$u\dfrac{d^2y}{du^2}+\dfrac{dy}{du}+4u^2(u^2-1)y=0$

Consider generally change the abscissa and the ordinate:

Let $u=f(v)$ ,

Then $\dfrac{dy}{du}=\dfrac{\dfrac{dy}{dv}}{\dfrac{du}{dv}}=\dfrac{1}{f'(v)}\dfrac{dy}{dv}$

$\dfrac{d^2y}{du^2}=\dfrac{d}{du}\left(\dfrac{1}{f'(v)}\dfrac{dy}{dv}\right)=\dfrac{\dfrac{d}{dv}\left(\dfrac{1}{f'(v)}\dfrac{dy}{dv}\right)}{\dfrac{du}{dv}}=\dfrac{\dfrac{1}{f'(v)}\dfrac{d^2y}{dv^2}-\dfrac{f''(v)}{(f'(v))^2}\dfrac{dy}{dv}}{f'(v)}=\dfrac{1}{(f'(v))^2}\dfrac{d^2y}{dv^2}-\dfrac{f''(v)}{(f'(v))^3}\dfrac{dy}{dv}$

$\therefore\dfrac{f(v)}{(f'(v))^2}\dfrac{d^2y}{dv^2}-\dfrac{f(v)f''(v)}{(f'(v))^3}\dfrac{dy}{dv}+\dfrac{1}{f'(v)}\dfrac{dy}{dv}+4(f(v))^2((f(v))^2-1)y=0$

$\dfrac{d^2y}{dv^2}+\left(\dfrac{f'(v)}{f(v)}-\dfrac{f''(v)}{f'(v)}\right)\dfrac{dy}{dv}+4f(v)((f(v))^2-1)(f'(v))^2y=0$

For example when take $f(v)=\dfrac{pv+q}{rv+s}$ , the ODE becomes

$\dfrac{d^2y}{dv^2}+\left(\dfrac{ps-qr}{(pv+q)(rv+s)}+\dfrac{2r}{rv+s}\right)\dfrac{dy}{dv}+\dfrac{4(ps-qr)^2(pv+q)((pv+q)^2-(rv+s)^2)}{(rv+s)^7}y=0$