Question

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

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

Answer 1

Answer: (A)

Given, $x^{2}ydx-(x^3+y^3)dy=0$
$\Rightarrow$ $\frac{dx}{dy}=\frac{x^3+y^3}{x^{2}y}$ ..(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)
$\Rightarrow$ $\frac{v^3}{3}=\log |y|-C$
$\Rightarrow$ $\log y=\frac{x^3}{3y^3}+C$
{from Eq. (ii)}