The solution of (x - y^3 )dx + 3x y^2 dy = 0 is - Vidyadip

MCQ

The solution of $(x - {y^3})dx + 3x{y^2}dy = 0$ is

A
$\log x + \frac{x}{y}$
✓
$\log x + \frac{{{y^3}}}{x} = k$
C
$\log x - \frac{x}{{{y^3}}} = k$
D
$\log xy - {y^3} = k$

✓

Answer

Correct option: B.

$\log x + \frac{{{y^3}}}{x} = k$

b
(b) $xdx - {y^3}dx + 3x{y^2}dy = 0$

Put ${y^3} = t$ $ \Rightarrow $ $dt = 3{y^2}dy$

$x\,dx - tdx + xdt = 0$ ==> $xdx + xdt - tdx = 0$

==> $\frac{{dx}}{x} + d\left( {\frac{t}{x}} \right) = 0$

On integration, we get $\log x + \frac{t}{x} = k$ ==> $\log x + \frac{{{y^3}}}{x} = k$.

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