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  • Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 31

Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 31

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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{d y}{d x}+P y=Q  where P,Q  are constants.

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} \\ & \end{aligned}

=\frac{d y}{d x}+P y=Q \\

P=\frac{2(x+2)}{x^{2}-1}, Q=\frac{2}{x-1}

If  of differential equation is

\begin{aligned} &I f=e^{\int \frac{2(x+2)}{x^{2}-1} d x} \\ & \end{aligned}

=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|} \\

=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}}}

\begin{aligned} &=e^{\ln (x+1)(x-1) \frac{(x-1)^{2}}{(x+1)^{2}}} \\ &=e^{\ln (x-1) \frac{(x-1)^{2}}{(x+1)}} \\ &=e^{\ln \frac{(x-1)^{3}}{(x+1)}} \\ &=\frac{(x-1)^{3}}{x+1} \\ &y \text { If }=\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 \\ & \end{aligned}

=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

\begin{aligned} &=2\left[\frac{t^{2}}{2}\right]+8 \log |t|-8 t+C \\ & \end{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

Posted by

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