Question

Solution of the equation $(x+2y^3)\frac{dy}{dx}-y=0$ is $(y>0)$

• A. $x=y^3+cy$
• B. $y=(1-xy)=cx$
• C. $x=(1-xy)=cy$
• D. $x=(1+xy)=cy$

Answer 1

Answer: (A)

$x=y^3+cy$

The following equation can be written as

$\frac{x+2y^3}{y}=\frac{dx}{dy}\therefore \frac{x}{y}+\frac{2y^3}{y}=\frac{dx}{dy}\frac{dx}{dy}-\frac{x}{y}=2y^2$

Which is similar to $\frac{dx}{dy}+Px=Q$

Where $P=-\frac{-1}{y}\&Q=2y^2$

Integrating factor$\left(IF\right)=e^{\int pdy}=e^{\int -\frac{1}{y}dy}=e^{log\frac{1}{y}}=\frac{1}{y}$

Solution is

$x.IF=\int Q.IFdyx.\frac{1}{y}=\int 2y^2\times \frac{1}{y}dy\frac{x}{y}=\int 2ydy\frac{x}{y}=\frac{2y^2}{2}+c\frac{x}{y}=y^2+cx=y^3+cy$

$\therefore$ Solution of equation $=y^3+cy$