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  • Provide solution for RD Sharma Maths Class12 Chapter 21 Differential Equation exercise 21.5 question 12.

Provide solution for RD Sharma Maths Class12 Chapter 21 Differential Equation exercise 21.5 question 12.

Answers (1)

Answer: y=\log \left | \log x \right |+\frac{x^2}{4}-\frac{x \sin2x}{4}-\frac{\cos 2x}{8}+C

Hint: You have to integrate by applying integration of xn.

Given: \frac{dy}{dx}-x\sin ^2x=\frac{1}{x\log x}

Solution: \frac{dy}{dx}=\frac{1}{x\log x}+x\sin ^2x

\begin{aligned} &\frac{d y}{d x}=\frac{1}{x \log x}+\frac{x}{2}(1-\cos 2 x) \\ &\frac{d y}{d x}=\frac{1}{x \log x}+\frac{x}{2}-\frac{x}{2} \cos 2 x \\ &d y=\left(\frac{1}{x \log x}+\frac{x}{2}-\frac{x}{2} \cos 2 x\right) d x \end{aligned}

Integrating both sides

\begin{aligned} &\int d y=\int \frac{1}{x \log x} d x+\frac{1}{2} \int x d x-\frac{1}{2} \int x \cos 2 x d x \\ &\left.y=\log |\log x|+\frac{1}{2} \frac{x^{2}}{2}-\frac{1}{2} \int x \times \cos 2 x d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \text { [ } \int f(x) g(x) d x=f(x) \int g(x) d x-\int\left(f^{\prime}(x) \int g(x) d x\right) d x\right] \end{aligned}

\begin{aligned} &y=\log |\log x|+\frac{x^{2}}{4}-\frac{x}{2} \int \cos 2 x d x+\frac{1}{2}\left[\int\left(\frac{d}{d x}(x) \int(\cos 2 x d x)\right) d x\right] \\ &y=\log |\log x|+\frac{x^{2}}{4}-\frac{x \sin 2 x}{4}-\frac{\cos 2 x}{8}+C \end{aligned}

Hence, y=\log \left | \log x \right |+\frac{x^2}{4}-\frac{x \sin2x}{4}-\frac{\cos 2x}{8}+C

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