How we can solve this :$y'(x)=\frac{y^3(x)}{y(x)+x}$?

Question

I have tried to get the solution of this ODE : $\displaystyle y'(x)=\frac{y^3(x)}{y(x)+x}$ using variable change $y=zx$ ,but i don't succeed to solve it , it's seems to me that it's solution should include error function . Then my question here is : How we can solve it ?

Answer 1

write your equation in the form $$-y(x)^3+(x+y(x))*y'(x)=0$$ now compute an integrating factor $$-y^3\frac{d \mu(y)}{dy}-3y^2\mu(y)=\mu(y)$$ and $$\frac{\frac{\partial\mu(y)}{\partial (y)}}{\mu(y)}=\frac{-3y^2-1}{y^3}$$ which gives $$\mu(y)=\frac{e^{\frac{1}{2y^2}}}{y^3}$$ and then $$-e^{\frac{1}{2y(x)^2}}+\frac{\left(e^{\frac{1}{2y(x)^2}}(x+y(x)\right)\frac{dy(x)}{dx}}{y(x)^3}=0$$ So fare on this evening. the rest Comes tomorrow.

Answer 2

$$y'(x)=\frac{y^3(x)}{y(x)+x}$$ Considering $x'$ rather than $y'$ $$y'(x)(y+x)=y^3$$ $$(y+x)=y^3x'$$ $$x'=\frac {(y+x)}{y^3}$$ $$x'-\frac {x}{y^3}=\frac 1 {y^2}$$ Then try to integrate $$\ln(x)=\int \frac {dy}{y^3}=\frac {-1}{2y^2}+K \implies x=Ke^{\frac {-1}{2y^2}}$$ $$K'e^{\frac {-1}{2y^2}}=\frac 1 {y^2} \implies K=C+\int \frac 1 {y^2} e^{\frac {1}{2y^2}} dy$$ $$x=Ce^{\frac {-1}{2y^2}}+e^{\frac {-1}{2y^2}}\int \frac {e^{\frac {1}{2y^2}}} {y^2} dy$$ The last integral is not easy to evaluate though $$........$$