Question

On solving the differential equation $ {{x}^{2}}y\,dx-({{x}^{3}}+{{y}^{2}})dy=0, $ the value of $ \log \,y $ is

• A. $ \frac{{{x}^{3}}}{3{{y}^{3}}}+C $
• B. $ \frac{{{x}^{2}}}{{{y}^{2}}}+C $
• C. $ \frac{{{x}^{2}}}{3{{y}^{3}}}+C $
• D. $ \frac{{{x}^{3}}}{{{x}^{3}}+{{y}^{3}}}+C $

Answer 1

Answer: (A)

Given, $ {{x}^{2}}y\,dx-({{x}^{3}}+{{y}^{3}})\,dy=0 $
$ \Rightarrow $ $ \frac{dx}{dy}=\frac{{{x}^{3}}+{{y}^{3}}}{{{x}^{2}}y} $ ..(i)
Which is a homogeneous differential equation Now, put $ x=vy $ ..(ii)
On differentiation wrt to y, we get
$ \frac{dx}{dy}=v.1+y\frac{dv}{dy} $
Then, from Eq. (i)
$ v+y\frac{dv}{dy}=\frac{{{(vy)}^{3}}+{{y}^{3}}}{{{(vy)}^{2}}y} $
$ =\frac{{{v}^{3}}+1}{{{v}^{2}}} $
$ \Rightarrow $ $ v+y\frac{dv}{dy}=v+\frac{1}{{{v}^{2}}} $
$ \Rightarrow $ $ y\frac{dv}{dy}=\frac{1}{{{v}^{2}}} $
$ \Rightarrow $ $ \int{{{v}^{2}}\,dv=\int{\frac{dy}{y}}} $
(on integrating)
$ \Rightarrow $ $ \frac{{{v}^{3}}}{3}=\log |y|-C $
$ \Rightarrow $ $ \log y=\frac{{{x}^{3}}}{3{{y}^{3}}}+C $
{from Eq. (ii)} $ $