Question

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

• A. $x=y^3+Cy$
• B. $x=y^2+Cy$
• C. $x^3=y^3+Cy$
• D. None of these.

Answer 1

The correct option is A

$x=y^3+Cy$

The explanation of the correct option:

Step-1 Forming linear differential equation:

The given differential equation: $\left(x+2y^3\right)\frac{dy}{dx}=y$.

$$\begin{gathered} \Rightarrow \frac{dy}{dx}=\frac{y}{\left(x+2y^3\right)} \\ \Rightarrow \frac{dx}{dy}=\frac{x+2y^3}{y} \\ \Rightarrow \frac{dx}{dy}=\frac{x}{y}+2y^2 \\ \Rightarrow \frac{dx}{dy}-\frac{x}{y}=2y^2 \\ \Rightarrow \frac{dx}{dy}-\frac{1}{y}x=2y^2 \end{gathered}$$

Compare the derived equation with $\frac{dx}{dy}+P\left(y\right)x=Q\left(y\right)$.

Thus, $P\left(y\right)=-\frac{1}{y}$ and $Q\left(y\right)=2y^2$.

Step-2: Integrating factor :

Thus, the integrating factor can be given by, $R=e^{\int P\left(y\right)dy}$

$$\begin{gathered} \Rightarrow R=e^{\int -\frac{1}{y}dy} \\ \Rightarrow R=e^{-\log y} \\ \Rightarrow R=\frac{1}{e^{\log y}} \\ \Rightarrow R=\frac{1}{y} \end{gathered}$$

Step-3 : Solution of linear of differential equation:

Consider the equation: $\frac{dx}{dy}-\frac{1}{y}x=2y^2$

Multiply both sides of the equation by the integrating factor.

$$\begin{gathered} \Rightarrow \frac{1}{y}\frac{dx}{dy}-\left(\frac{1}{y}\times \frac{1}{y}x\right)=2y^2\times \frac{1}{y} \\ \Rightarrow \frac{1}{y}\frac{dx}{dy}-\frac{x}{y^2}=2y \\ \Rightarrow \frac{d}{dy}\left(\frac{x}{y}\right)=2y \\ \Rightarrow d\left(\frac{x}{y}\right)=2ydy \end{gathered}$$

Integrate both sides of the equation.

$$\begin{gathered} \Rightarrow \int d\left(\frac{x}{y}\right)=\int 2ydy \\ \Rightarrow \frac{x}{y}=2\times \frac{y^2}{2}+C \\ \Rightarrow \frac{x}{y}=y^2+C \\ \Rightarrow x=y^3+Cy \end{gathered}$$

Therefore, the solution of the differential equation $\left(x+2y^3\right)\frac{dy}{dx}=y$ is $x=y^3+Cy$.

Hence, option A is correct.