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

Question
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, $$int e^t dt= int dx/x+c$$
$$e^t= lnx+c$$
$$t= ln(lnx+c)$$ (use property $$e^x=y => x=ln y$$)
Substitute back $$t=v^2$$
$$v^2=ln(lnx+c)$$
Substitute back $$v=y/x$$
$$(y/x)^2= ln(lnx+c)$$
$$y^2/x= ln(lnx+c)$$
$$y^2= x^2 ln(lnx+c)$$
$$y= +- sqrt(x^2 ln(lnx+c))$$
$$y= +- x sqrt(ln(lnx+c)) ### Relevant Questions asked 2021-01-16 Find the general solution of the first-order linear differential equation \((dy/dx)+(1/x)y = 6x + 2$$, for x > 0
Solve differential equation $$\displaystyle{2}{x}{y}-{9}{x}^{{2}}+{\left({2}{y}+{x}^{{2}}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={0},\ {y}{\left({0}\right)}=-{3}$$
Solve differential equation $$2xy-9x^2+(2y+x^2+1)dy/dx=0$$, y(0)= -3
Solve differential equation $$1+y^2+xy^2)dx+(x^2y+y+2xy)dy=0$$
Solve differential equation $$((3y^2-t^2)/y^5)dy/dx+t/(2y^4)=0$$, y(1)=1
Solve differential equation $$(6x+1)y^2 dy/dx+3x^2+2y^3=0$$, y(0)=1
Solve differential equation $$\displaystyle{x}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}-{2}{y}={x}^{{3}}{{\sin}^{{2}}{\left({x}\right)}}$$
Solve differential equation $$(cos^2y)/(4x+2)dy= ((cosy+siny)^2)/(sqrt(x^2+x+3))dx$$
Solve differential equation $$\displaystyle{y}{\left({2}{x}-{2}+{x}{y}+{1}\right)}{\left.{d}{x}\right.}+{\left({x}-{y}\right)}{\left.{d}{y}\right.}={0}$$
Solve differential equation $$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\left({x}+{y}+{1}\right)}^{{2}}-{\left({x}+{y}-{1}\right)}^{{2}}$$