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  • Explain solution RD Sharma Class 12 Chapter 21 Differential Equation exercise 21.5 question 19 Maths.

Explain solution RD Sharma Class 12 Chapter 21 Differential Equation exercise 21.5 question 19 Maths.

Answers (1)

Answer:  y=\frac{x^2}{2}\log x-\frac{1}{2}\frac{x^2}{2}+C

Hint: You have to integrate by applying integration of \log x.

Given: \frac{dy}{dx}=x \log x

Solution: \int dy=\int x \log x\: dx

Integrate on both sides

\begin{aligned} &=\int d y=\log x \cdot \frac{x^{2}}{2}-\int\left(\frac{1}{x} \cdot \frac{x^{2}}{2} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\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]\right. \\ &=y=\frac{x^{2}}{2} \log x-\int \frac{x}{2} d x \\ &=y=\frac{x^{2}}{2} \log x-\frac{1}{2} \int x d x \\ &=y=\frac{x^{2}}{2} \log x-\frac{1}{2} \frac{x^{2}}{2}+C \end{aligned}

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