Question

The Solution of the differential equation $\left(x+2y^{3}\right) \frac{dy}{dx}=y$ is

• A. $y^{3}+Cy=x$
• B. $x+2y^{3}=y+C$
• C. $y^{3}+Cx=y$
• D. $\frac{xy^{4}}{2}+xy=Cy$

Answer 1

Answer: (A)

$y^{3}+Cy=x$

Answer 2

Given, $\left(x+2 y^{3}\right) \frac{d y}{d x}=y$
$\Rightarrow y \frac{d x}{d y}=x+2 y^{3}$
$\Rightarrow \frac{d x}{d y}-\frac{1}{y} x=2 y^{2}$
This is of the form $\frac{d x}{d y}+P x=Q$.
where, $P=-\frac{1}{y}$ and $Q=2 y^{2}$
Thus, the given equation is linear.
$\therefore IF = e ^{\int Pdy }= e ^{\int-\frac{1}{y} d y}= e ^{-\log y}= e ^{\log (y)^{-1}}= y ^{-1}=\frac{1}{ y }$
So, the required solution is
$x \cdot I F =\int(Q \cdot \mid F) d y +C$
$\Rightarrow x \cdot \frac{1}{y} =\int\left(2 y^{2} \cdot \frac{1}{y}\right) d y +C$
$\Rightarrow x \cdot \frac{1}{y} =\int 2 y d y +C=y^{2}+C$
$\Rightarrow x =y^{3}+ C y$