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  • Explain solution RD Sharma Class 12 Chapter 21 Differential Equation exercise 21.5 question 23 Maths.

Explain solution RD Sharma Class 12 Chapter 21 Differential Equation exercise 21.5 question 23 Maths.

Answers (1)

Answer: C=3

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

Given: e^{\frac{dy}{dx}}=x+1,y(0)=3

Solution: \frac{dy}{dx}=\ln (x+1)dx

             dy=\ln (x+1)dx

Partial integration on both sides

\begin{aligned} &\int d y=\int 1 \ln (x+1) d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\int f(x) g(x) d x=f(x) \int g(x) d x-\int f^{\prime}(x) \int g(x) d x\right] \\ &y=x \ln (x+1)-\int \frac{x}{x+1} d x \\ &y=x \ln (x+1)-\int \frac{x+1-1}{x+1} d x \\ &y=x \ln (x+1)-\int \frac{x+1-1}{x+1} d x \\ &y=x \ln (x+1)-\int 1-\frac{1}{x+1} d x \end{aligned}

\begin{aligned} &y=x \ln (x+1)-x \cdot \ln |x+1|+C \\ &y=x \ln (x+1)-x+\ln (x+1)+C \end{aligned}

Put x=0, y=3 we get

\begin{aligned} &y=x \ln (x+1)-x+\ln (x+1)+C\\ &3=0 \ln (0+1)-0+\ln (0+1)+C\\ &3=0+C\\ &C=3 \end{aligned}

 

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