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  • Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 31 textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 31 textbook solution.

Answers (1)

Answer : \frac{y(x-1)^{3}}{x+1}\left(x^{2}-6 x+8 \log |x+1|\right)+C

Hint : To solve this equation we use \frac{dy}{dx}+Py=Q where P,Q are constnats.

Give : \left(x^{2}-1\right) \frac{d y}{d x}+2(x+2) y=2(x+1)

Solution :

\begin{aligned} &\frac{d y}{d x}+\frac{2(x+2)}{x^{2}-1} y=\frac{2}{x-1} \\ &=\frac{d y}{d x}+P y=Q \\ &P=\frac{2(x+2)}{x^{2}-1}, Q=\frac{2}{x-1} \end{aligned}

I\, f of differential equation is

\begin{aligned} &I f=e^{\int \frac{2(x+2)}{x^{2}-1} d x} \\ &=e^{\int\left(\frac{2 x}{x^{2}-1}+\frac{4}{x^{2}-1}\right) d x} \\ &=e^{\int \ln \left|x^{2}-1\right|+4 \times \frac{1}{2} \ln \left|\frac{x-1}{x+1}\right|} \\ &=e^{\int \ln \left|x^{2}-1\right|+2 \ln \left|\frac{x-1}{x+1}\right|} \end{aligned}

\begin{aligned} &=e^{\int \ln \left|x^{2}-1\right|+\ln \left|\frac{(x-1)^{2}}{(x+1)^{2}}\right|} \\ &=e^{\ln \left(x^{2}-1\right) \frac{(x-1)^{2}}{(x+1)^{2}}} \\ &=e^{\ln (x+1)(x-1) \frac{(x-1)^{2}}{(x+1)^{2}}} \\ &=e^{\ln (x-1) \frac{(x-1)^{2}}{(x+1)}} \end{aligned}

\begin{aligned} &=e^{\ln \frac{(x-1)^{3}}{(x+1)}} \\ &=\frac{(x-1)^{3}}{x+1} \\ &y I f=\int \text { QIf } d x+C \end{aligned}

\begin{aligned} &=y \frac{(x-1)^{3}}{x+1}=\int\left(\frac{2}{x-1}\right) \frac{(x-1)^{3}}{x+1} d x+C \\ &=2 \int \frac{(x-1)^{3}}{x+1} d x+C \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[x+1=t, d x=d t] \\ &=2 \int \frac{\left(t^{2}+4-4 t\right)}{t} d t \\ &=2\left[\frac{t^{2}}{2}\right]+8 \log |t|-8 t+C \end{aligned}

\begin{aligned} &=t^{2}+8 \log |t|-8 t+C \\ &=(x+1)^{2}+8 \log |x+1|-8(x+1)+C \\ &=x^{2}+2 x+1+8 \log |x+1|-8 x-8+C \\ &=y \frac{(x-1)^{3}}{(x+1)}\left(x^{2}-6 x+8 \log |x+1|\right)+C \end{aligned}

 

 

Posted by

infoexpert23

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