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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

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{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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