Solutions of the differential equation of the form: $y=y'-\frac{x^{-1/2}}{n}$

Question

I was experimenting with the expansion of "e" and discovered this expansion:

Let $y= \frac{x^{1/2}}{Γ(3/2)}+\frac{x^{3/2}}{Γ(5/2)}+\frac{x^{5/2}}{Γ(7/2)}+...$

This function upon differentiation gives:

$y'=\frac{x^{-1/2}}{Γ(1/2)}+\frac{x^{1/2}}{Γ(3/2)}+\frac{x^{3/2}}{Γ(5/2)}+...$

Which is essentially y plus the first term ($\frac{x^{-1/2}}{Γ(1/2)}$) from y'

Therefore $y=y'-\frac{x^{-1/2}}{\sqrt{\pi}}$

However I cannot reduce this expression further and have seem to hit a roadblock. Is the solution for this equation even derivable?

Answer 1

Hint This equation is first order and linear. Rearranging gives $$y' - y = \frac{1}{\sqrt{\pi} \sqrt{x}},$$ so our integrating factor is $$\exp \int_{x_0} -dx = e^{-x},$$ hence we write $$e^{-x} y' - e^{-x} y = \frac{1}{\sqrt{\pi}} \frac{e^{-x}}{\sqrt{x}} .$$ Now, the left-hand side is a derivative of an expression (which?) in $x, y$. To integrate the right-hand side you'll need the error function, $\operatorname{erf}$, or the equivalent to express your answer.

Integrating and rearranging gives

$$y(x) = e^x (\operatorname{erf}\sqrt x + C).$$ The Ratio Test shows that the series converges for all $x \geq 0$, and $0 = \lim_{x \searrow 0} y(x) = C$, so $$y(x) = e^x \operatorname{erf}\sqrt x.$$

Answer 2

Because of the solution of the homegeous equation, let $y=z \,e^x$ to make $$z'=\frac{e^{-x}}{n \sqrt{x}}\implies z=C+\frac 1 n\int \frac{e^{-x}}{ \sqrt{x}}\,dx$$ Let $x=t^2$ $$\int \frac{e^{-x}}{ \sqrt{x}}\,dx=2 \int e^{-t^2}\,dt=\sqrt{\pi }\,\text{erf}(t)=\sqrt{\pi } \,\text{erf}\left(\sqrt{x}\right)$$ So $$y=e^x \left(C+ \frac{\sqrt{\pi }}n \,\text{erf}\left(\sqrt{x}\right) \right)$$