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Please solve RD Sharma Class 12 Chapter 21 Differential Equation Exercise Revision Exercise (RE) Question 54 maths textbook solution.

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Answer : \tan \frac{y}{x}=-\log |x|+c

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

Given :     x \frac{d y}{d x}+x \cos ^{2}\left(\frac{y}{x}\right)=y

Solution : x \frac{d y}{d x}+x \cos ^{2}\left(\frac{y}{x}\right)=y

                  \begin{aligned} &\frac{d y}{d x}+\cos ^{2}\left(\frac{y}{x}\right)=\frac{y}{x} \\ &\frac{d y}{d x}=\frac{y}{x}-\cos ^{2}\left(\frac{y}{x}\right) \end{aligned}

Putting y=vx and differentiate

Therefore, \frac{d y}{d x}=v+x \frac{d v}{d x}


 \begin{aligned} &v+x \frac{d v}{d x}=v-\cos ^{2}(v) \\ &\frac{1}{\cos ^{2} v} d v=-\frac{d x}{x} \\ &\sec ^{2} v d v=-\frac{1}{x} d x \end{aligned}

Integrating both sides,

 \begin{aligned} \int \sec ^{2} d v &=-\int \frac{1}{x} d x \\ \tan v &=-\log |x|+c \\ \tan \frac{y}{x} &=-\log |x|+c \end{aligned}

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