Q. To reduce the differential equation $\frac{dy}{dx} + P(x).y = Q(x).y^{n}$ 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) $v = y^{n - 1}$
Explanation
(b)
The given differential equation is
$\frac{dy}{dx} + P(x)y = Q(x).y^{n}$
$\Rightarrow$ $\frac{1}{y^{n}}.\frac{dy}{dx} + y^{- n + 1}P(x) = Q(x)$
Put $\frac{1}{y^{n - 1}} = v$
$\Rightarrow$ $( - n + 1)y^{- n}\ \frac{dy}{dx} = \frac{dv}{dx}$
$\therefore$ $\frac{1}{( - n + 1)}.\frac{dv}{dx} + P(x).v = Q(x)$
$\Rightarrow$ $\frac{dv}{dx} + (1 - n)P(x)v = (1 - n)Q(x)$
Hence, required substitution is $v = \frac{1}{y^{n - 1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.