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प्रश्न
Find the particular solution of the differential equation `dy/dx=(xy)/(x^2+y^2)` given that y = 1, when x = 0.
उत्तर
`dy/dx=(xy)/(x^2+y^2) .....(1)`
This is a homogenous differential equation.
Substitute y = vx .....(2)
`⇒dy/dx=v+x (dv)/dx .....(3)`
From (1), (2) and (3), we have
`x (dv)/dx+v=(x (vx))/(x^2+(vx)^2)=(vx^2)/(x^2 (1+v^2))`
`⇒x (dv)/dx+v=v/(1+v^2)`
`⇒x (dv)/dx=v/(1+v^2)-v=(v-v-v^2)/(1+v^2)`
`⇒x (dv)/dx=−v^3/(1+v^2)`
`⇒(1+v^2)/v^3dv=−dx/x`
`⇒(1/v^3+1/v)dv=−dx/x`
Integrating both sides, we have
`v^(−3+1)/(−3+1)+lnv=−lnx+C`
`⇒−1/(2v^2)+lnv=−lnx+C`
`⇒−1/(2v^2)+lnvx=C`
`⇒−x^2/(2y^2)+lny=C`
Given: y = 1 when x = 0
⇒ C = 0
Thus, the particular solution of the given differential equation is given by
`lny=x^2/(2y^2)`
or x2 = 2y2 lny
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