Question

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

First order differential equations
ANSWERED
asked 2021-02-21
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, \(\int e^t dt= \int dx/x+c\)
\(e^t= \ln x+c\)
\(t= \ln(\ln x+c)\) (use property \(e^x=y => x=\ln y\))
Substitute back \(t=v^2\)
\(v^2=\ln(\ln x+c)\)
Substitute back \(v=y/x\)
\((y/x)^2= \ln(\ln x+c)\)
\(y^2/x= \ln(\ln x+c)\)
\(y^2= x^2 \ln(\ln x+c)\)
\(y= +- \sqrt{x^2 \ln(\ln x+c)}\)
\(y= \pm x\sqrt{\ln (\ln x+C)}\)

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