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

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

Answers (1)

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}

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