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  • Explain solution RD Sharma Class 12 Chapter 21 Differential Equation exercise 21.5 question 26 Maths.

Explain solution RD Sharma Class 12 Chapter 21 Differential Equation exercise 21.5 question 26 Maths.

Answers (1)

Answer: C=\ln \left |2 \right |-\frac{1}{2}\ln \left |3 \right |

Hint: Find the value of C.

Given: x(x^2-1)\frac{dy}{dx}=1, y(2)=0

Solution: \frac{dy}{dx}=\frac{1}{ x(x^2-1)}

dy=\frac{dx}{ x(x^2-1)}

Integrate on both sides

\begin{array}{r} \int d y=\int \frac{d x}{x\left(x^{2}-1\right)} \\ \int d y=\int \frac{d x}{x(x-1)(x+1)} \end{array}

\begin {gathered} \frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1} \\ \frac{\frac{1}{(-1)(1)}}{x}+\frac{1}{(1)(2)}{x-1}+\frac{\frac{1}{(-1)(-2)}}{x-2} \\ \int-\frac{1}{x}+\int \frac{1}{2(x-1)}+\int \frac{1}{2(x-2)} \\ =y=\ln |x|+\frac{1}{2} \ln |x-1|+\frac{1}{2} \ln |x+1|+C \end{gathered}

put values, y(2) = 0

\begin{aligned} &0=-\ln |2|+\frac{1}{2} \ln |1|+\frac{1}{2} \ln |3|+C \\ &C=\ln |2|-\frac{1}{2} \ln |3| \end{aligned}

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