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

UkusakazaL 2021-02-21 Answered
Find general solution of the following differential equation dy/dx=(2y2+x2e((y/x)2))/(2xy)
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Expert Answer

dieseisB
Answered 2021-02-22 Author has 85 answers

y=vx=>y/x=v
dy/dx=v+x(dv)/dx
v+x(dv)/dx=(2(vx)2+x2e((v)2))/(2x(vx))
v+x(dv)/dx=(2v2x2+x2e((v)2))/(2vx2)
v+x(dv)/dx=(x2(2v2+e((v)2)))/(2vx2) (take x2 as common factor from numerator)
v+x(dv)/dx=(2v2+e((v)2))/(2v) (divide x2/x2=1)
Substract v from both sides and further simplify it
v+x(dv)/dxv=(2v2+e((v)2))/(2v)v
x(dv)/dx=(2v2e((v)2)2v2)/(2v)
x(dv)/dx=e((v)2)/(2x)
(2v)/e((v)2)dv=dx/x
Integrate both sides
2vev2dv=dx/x+c
Substitute v2=t
=>2vdv=dt
Hence, etdt=dx/x+c
et=lnx+c
t=ln(lnx+c) (use property ex=y=>x=lny)
Substitute back t=v2
v2=ln(lnx+c)
Substitute back v=y/x
(y/x)2=ln(lnx+c)
y2/x=ln(lnx+c)
y2=x2ln(lnx+c)
y=+x2ln(lnx+c)
y=±xln(lnx+C)

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Jeffrey Jordon
Answered 2021-12-12 Author has 2313 answers

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