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Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 64 Subquestion (iv) textbook solution.

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Answer : y=e^{cx}

Hint              : You must know the rules of solving differential equation and integration

Given            :    y \log y d x-x d y=0

Solution         : y \log y d x-x d y=0

                         y \log y d x=x\; dy

                           \begin{aligned} \frac{1}{x} d x &=\frac{1}{y \log y} d y \\ \frac{1}{y \log y} d y &=\frac{1}{x} d x \end{aligned}

Integrating both sides,

                 \int \frac{1}{y \log y} d y=\int \frac{d x}{x}

Put log\; y =t and differentiating,



\begin{aligned} &\int \frac{1}{t} d t=\int \frac{1}{x} d x \\ &\log |t|=\log x+\log c \\ &\log (\log y)=\log x+\log c \end{aligned}

\begin{aligned} &\log (\log y)=\log c x \\ &\log y=c x \\ &y=e^{c x} \end{aligned}

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