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  • Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 24 maths textbook solution.

Explain solution for  RD Sharma Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 24 maths textbook solution.

Answers (1)

Answer : (c) u=(\log z)^{-1}

Hint: Divide the differentiated equation by z(\log z)^{2} to make the equation linear in u

Given: \frac{d z}{d x}+\frac{z}{x} \log z=\frac{z}{x^{2}}(\log z)^{2}

Explanation: Divide both sides by  z(\log z)^{2}

\begin{aligned} &\Rightarrow \frac{1}{z(\log z)^{2}} \frac{d z}{d x}+\frac{\frac{z \log z}{x}}{z(\log z)^{2}}=\frac{\frac{z(\log z)^{2}}{x^{2}}}{z(\log z)^{2}} \\ &\Rightarrow \frac{1}{z(\log z)^{2}} \frac{d z}{d x}+\frac{1}{x \log z}=\frac{1}{x^{2}} \end{aligned}   .....(i)

Now, compare it with

\begin{aligned} &\Rightarrow \frac{d u}{d x}+u P(x)=Q(x) \\ &\text { Then let } u=(\log z)^{-1} \end{aligned}                                   ....(ii)

\begin{aligned} &\Rightarrow \frac{d u}{d x}=\frac{-1}{(\log z)^{2}} \frac{1}{z} \frac{d z}{d x} \\ &\Rightarrow \frac{d u}{d x}=\frac{-1}{z(\log z)^{2}} \frac{d z}{d x} \end{aligned}

So the above equation is of the form

\begin{aligned} &\Rightarrow \frac{d u}{d x}+u P(x)=Q(x) \\ &\text { Where } P(x)=-\frac{1}{x} \text { and } Q(x)=-\frac{1}{x^{2}} \end{aligned}

Hence,

\Rightarrow u=(\log z)^{-1}                      [from (ii)]

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