Question

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

• A. $y=C\left(x+a\right)\left(1-ay\right)$
• B. $y=C\left(x+a\right)\left(1+ay\right)$
• C. $y=C\left(x-a\right)\left(1+ay\right)$
• D. None of these.

Answer 1

The correct option is A

$y=C\left(x+a\right)\left(1-ay\right)$

The explanation for the correct option:

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

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

Integrate both sides of the equation.

$$\begin{gathered} \Rightarrow \int \left[\frac{dy}{y}+\frac{ady}{\left(1-ay\right)}\right]=\int \frac{dx}{\left(x+a\right)} \\ \Rightarrow \int \frac{dy}{y}+a\int \frac{dy}{\left(1-ay\right)}=\log \left|x+a\right|+C_1\left[{\mathbf{C}}_{\mathbf{1}}\mathbf{=}\mathbf{Integrating}\mathbf{}\mathbf{constant}\right] \\ \Rightarrow \log \left|y\right|+\frac{a}{\left(-a\right)}\log \left|1-ay\right|=\log \left|x+a\right|+C_1 \\ \Rightarrow \log \left|y\right|-\log \left|1-ay\right|=\log \left|x+a\right|+C_1 \\ \Rightarrow \log \left|\frac{y}{1-ay}\right|=\log \left|x+a\right|+\log C\left[C_1=\mathrm{logC}\right] \\ \Rightarrow \log \left|\frac{y}{1-ay}\right|=\log C\left|x+a\right| \\ \Rightarrow e^{\log \left|\frac{y}{1-ay}\right|}=e^{\log C\left|x+a\right|} \\ \Rightarrow \frac{y}{1-ay}=C\left(x+a\right) \\ \Rightarrow y=C\left(x+a\right)\left(1-ay\right) \end{gathered}$$

Therefore, the solution of the equation $y-x\frac{dy}{dx}=a\left(y^2+\frac{dy}{dx}\right)$ is $y=C\left(x+a\right)\left(1-ay\right)$.

Hence, option A is correct.