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 \(dy/dx=(2y^2+x^2e^(-(y/x)^2))/(2xy)\)

Answers (1)

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))
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