Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise Multiple choice Question Question 36 maths textbook solution.

Answer : $(d) z=y^{1-n}$

Hint: Divide both sides  $y^{n}$

Given : $\frac{d y}{d x}+P y=Q y^{n}, n>2$

Explanation: Divide both sides  $y^{n}$

\begin{aligned} &\Rightarrow y^{-n} \frac{d y}{d x}+\frac{P y}{y^{n}}=\frac{Q y^{n}}{y^{n}} \\ &\Rightarrow y^{-n} \frac{d y}{d x}+y^{1-n} P=Q \end{aligned}                                  ....(i)

Let $y^{1-n}=z$

\begin{aligned} &\Rightarrow(1-n) y^{1-n-1} \frac{d y}{d x}=\frac{d z}{d x} \\ &\Rightarrow(1-n) y^{-n} \frac{d y}{d x}=\frac{d z}{d x} \end{aligned}

$\Rightarrow y^{-n} \frac{d y}{d x}=\frac{1}{1-n} \frac{d z}{d x}$

Put the value in (i) we get

\begin{aligned} &\Rightarrow \frac{1}{1-n} \frac{d z}{d x}+y^{1-n} P=Q \\ &\Rightarrow \frac{1}{1-n} \frac{d z}{d x}+z P=Q \end{aligned}

\begin{aligned} &\Rightarrow \frac{d z}{d x}+(1-n) z P=Q(1-n) \\ &\text { As } n>2, \text { so }(1-n) \in z \end{aligned}

Given differential equation is linear

Hence, $z=y^{1-n}$