How can I solve $y''-2xy'=-2$?

Question

How can I solve this differential equation ? I search a lot about the answer but i can not find it. $$\frac12y''-xy'=-1\qquad u=y'$$ $$\frac12u'-xu=-1\qquad u=ue^{x^2}$$ $$y=\int ue^{x^2}\,dx$$ I don't know how can i solve last integral. thanks for any help.

Answer 1

Let $z:=y'$. Then, you have the differential equation $$z'(x)-2x\,z(x)=-2\,.$$ Therefore, with $u(x):=\exp(-x^2)\,z(x)$, we have $$u'(x)=\exp(-x^2)\,\big(z'(x)-2x\,z(x)\big)=-2\,\exp(-x^2)\,.$$ Thus, $$u(x)=\int_0^x\,\big(-2\,\exp(-t^2)\big)\,\text{d}t+\frac{2}{\sqrt{\pi}}\,a=-\sqrt{\pi}\,\text{erf}(x)+\frac{2}{\sqrt{\pi}}\,a$$ for some constant $a$. (The function $\text{erf}$ is the error function.)

Therefore, $$y'(x)=z(x)=\exp(x^2)\,u(x)=-\sqrt{\pi}\,\text{erf}(x)\,\exp(x^2)+\frac{2}{\sqrt{\pi}}\,a\,\exp(x^2)\,.$$ Ergo, $$y(x)=-\sqrt{\pi}\,\int_0^x\,\text{erf}(t)\,\exp(t^2)\,\text{d}t+a\,\text{erfi}(x)+b$$ for some constant $b$. Here, $\text{erfi}$ is the imaginary error function. Note that $y$ can be written in the form of generalized hypergeometric functions: $$y(x)=-x^2\,{_2\text{F}_2}\left(1,1;\frac32,2;x^2\right)+a\,\text{erfi}(x)+b\,.$$