Question

If the differential equation $3x^{\frac{1}{3}}dy+x^{- \frac{2}{3}}ydx=3xdx$ is satisfied by $kx^{\frac{1}{3}}y=x^{2}+c$ (where $c$ is an arbitrary constant), then the value of $k$ is

• A. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

The given equation is $x^{\frac{1}{3}}.dy+\frac{1}{3}x^{- \frac{2}{3}}dx.y=xdx$
or $d\left(x^{\frac{1}{3}} . y\right)=xdx$
Integrating, we get,
$x^{\frac{1}{3}}.y=\frac{x^{2}}{2}+\lambda $
Or $2x^{\frac{1}{3}}y=x^{2}+C$
$\Rightarrow k=2$