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

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)]