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

Answer: (B)

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

An equation of the form $\frac{dy}{dx}+Py=Q{u}^n$ where P and Qare functions ofx alone or constants, is called Bernoulli's equation. Divide both the sides by y${}^n$, we get
$y^{-n}\frac{dy}{dx}+P{y}^{-n+1}=Q$ Put
$y^{-n+1}=z\Rightarrow (-n+1)y^{-n}\frac{dy}{dx}=\frac{dz}{dx}$.
The equation reduces to $\frac{1}{1-n}\frac{dz}{dx}+Pz=Q\Rightarrow \frac{dz}{dx}+(1-n)Pz=Q$
Which is linear in z and can be solved in the usual manner.
So the substitution is $z=v=\frac{1}{y^{n-1}}$