If the differential equation 3x(1/3)dy+x- (2/3)ydx=3xdx is satisfied by kx(1/3)y=x2+c (where c is an arbitrary constant), then the value of k is

Question

If the differential equation 3x(1/3)dy+x- (2/3)ydx=3xdx is satisfied by kx(1/3)y=x2+c (where c is an arbitrary constant), then the value of k is (A) (1/3) (B) (2/3) (C) 2 (D) 1

Tardigrade Admin · Accepted Answer

The given equation is x(1/3).dy+(1/3)x- (2/3)dx.y=xdx or d(x(1/3) . y)=xdx Integrating, we get, x(1/3).y=(x2/2)+λ Or 2x(1/3)y=x2+C ⇒ k=2

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

$\frac{1}{3}$

$\frac{2}{3}$

$2$

$1$

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

Q. If the differential equation 3x31​dy+x−32​ydx=3xdx is satisfied by kx31​y=x2+c (where c is an arbitrary constant), then the value of k is

31​

32​

2

1

The given equation is x31​.dy+31​x−32​dx.y=xdx or d(x31​.y)=xdx Integrating, we get, x31​.y=2x2​+λ Or 2x31​y=x2+C ⇒k=2

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

$\frac{1}{3}$

$\frac{2}{3}$

$2$

$1$

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