Question

# Find general solution of the following differential equation dy/dx=(2y^2+x^2e^(-(y/x)^2))/(2xy)

First order differential equations
Find general solution of the following differential equation $$dy/dx=(2y^2+x^2e^(-(y/x)^2))/(2xy)$$

2021-02-22

$$y=vx => y/x=v$$
$$dy/dx= v+x (dv)/dx$$
$$v+x (dv)/dx= (2(vx)^2+x^2e^(-(v)^2))/(2x(vx))$$
$$v+x (dv)/dx= (2v^2x^2+x^2e^(-(v)^2))/(2vx^2)$$
$$v+x (dv)/dx= (x^2(2v^2+e^(-(v)^2)))/(2vx^2)$$ (take $$x^2$$ as common factor from numerator)
$$v+x (dv)/dx=(2v^2+e^(-(v)^2))/(2v)$$ (divide $$x^2/x^2=1$$)
Substract v from both sides and further simplify it
$$v+x (dv)/dx-v= (2v^2+e^(-(v)^2))/(2v)-v$$
$$x(dv)/dx= (2v^2e^(-(v)^2)-2v^2)/(2v)$$
$$x (dv)/dx= e^(-(v)^2)/(2x)$$
$$(2v)/e^(-(v)^2) dv= dx/x$$
Integrate both sides
$$\int 2ve^{v^2}dv= \int dx/x+c$$
Substitute $$v^2=t$$
$$=> 2vdv=dt$$
Hence, $$\int e^t dt= \int dx/x+c$$
$$e^t= \ln x+c$$
$$t= \ln(\ln x+c)$$ (use property $$e^x=y => x=\ln y$$)
Substitute back $$t=v^2$$
$$v^2=\ln(\ln x+c)$$
Substitute back $$v=y/x$$
$$(y/x)^2= \ln(\ln x+c)$$
$$y^2/x= \ln(\ln x+c)$$
$$y^2= x^2 \ln(\ln x+c)$$
$$y= +- \sqrt{x^2 \ln(\ln x+c)}$$
$$y= \pm x\sqrt{\ln (\ln x+C)}$$