Question

To reduce the differential equation $\frac{dy}{dx}+P\left(x\right)·y=Q\left(x\right)·y^n$ to the linear form, the substitution is

• A. $v=\frac{1}{y^n}$
• B. $v=\frac{1}{y^{n-1}}$
• C. $v=y^n$
• D. $v=y^{n-1}$

Answer 1

The correct option is B

$v=\frac{1}{y^{n-1}}$

Explanation for the correct option:

Find the substitution to reduce the given differential equation in the linear differential equation.

In the question, an equation $\frac{dy}{dx}+P\left(x\right)·y=Q\left(x\right)·y^n$ is given. This is also known as Bernoulli's equation.

Divide both sides of the given differential equation by $y^n$.

$$\begin{gathered} \frac{d\left(y^{-n+1}\right)}{dx}+P\left(x\right)·y^{-n+1}=Q\left(x\right) \\ \Rightarrow (-n+1)y^{-n}\frac{dy}{dx}+P\left(x\right)·y^{-n+1}=Q\left(x\right)...1 \end{gathered}$$

Assume that, $v=y^{-n+1}$.

Differentiate both sides with respect to $x$.

$\frac{dv}{dx}=(-n+1)y^{-n}\frac{dy}{dx}$

So, equation $1$ becomes $\frac{dv}{dx}+P\left(x\right)·v=Q\left(x\right)$ which is a linear differential equation.

Therefore, to reduce the differential equation $\frac{dy}{dx}+P\left(x\right)·y=Q\left(x\right)·y^n$ to the linear form, the substitution is $v=\frac{1}{y^{n-1}}$.

Hence, option $B$ is the correct option.