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Question
Show that the differential equation `2xydy/dx=x^2+3y^2` is homogeneous and solve it.
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Solution
The given differential equation can be expressed as
`dy/dx=(x^2+3y^2)/(2xy) .....(i)`
`Let F(x, y)=(x^2+3y^2)/(2xy)`
Now,
`F(λx, λy)=((λx)^2+3(λy)^2)/(2(λx)(λy))=(λ^2(x^2+3y^2))/(λ^2(2xy))=λ^0F(x, y)`
Therefore, F(x, y) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.
Let y = vx .....(ii)
Differentiating (ii) w.r.t. x, we get
`dy/dx=v+x(dv)/dx`
Substituting the value of y and dy/dx in (i), we get
`v+x(dv)/dx=(1+3v^2)/(2v)`
`⇒x(dv)/dx=(1+3v^2)/(2v)−v`
` ⇒x(dv)/dx=(1+3v^2−2v^2)/(2v)`
`⇒x(dv)/dx=(1+v^2)/(2v)`
`⇒(2v)/(1+v^2)dv=dx/x .....(ii)`
Integrating both side of (iii), we get
`∫(2v)/(1+v^2)dv=∫dx/x`
Putting `1+v^2=t`
⇒2vdv=dt
`∴∫dt/t=∫dx/x`
⇒log|t|=log|x| +log|C1|
⇒log∣t/x∣=log|C1|
`⇒t/x=±C_1`
`⇒(1+v^2)/x=±C_1`
`⇒(1+y^2/x^2)/x=±C_1`
x2+y2=Cx3
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