#### Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 20 maths

Answer:  $y=\frac{x}{x+1}(x+\log x-1)$

Given: $x(x+1) \frac{d y}{d x}-y=x(x+1)$

To find: We have to show that the curve which satisfies $x(x+1) \frac{d y}{d x}-y=x(x+1)$  and passes through $\left ( 1,0 \right )$

Hint: Use linear differential equation to solve.

Solution: $x(x+1) \frac{d y}{d x}-y=x(x+1)$

Dividing by  $x(x+1)$

$=\frac{d y}{d x}-\frac{y}{x(x+1)}=1$        [This is a linear differential equation]

Comparing with $\frac{d y}{d x}+P y=Q$   we get

$P=\frac{1}{x(x+1)}, Q=1$

$\text { Now If }=e^{\int P d x}$

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

\begin{aligned} &=e^{-\log \left(\frac{x}{x+1}\right)} \\\\ &=e^{\log \left(\frac{x}{x+1}\right)^{-1}} \quad\quad\quad\left[e^{-\log x}=\log x^{-1}\right] \end{aligned}

$=\left(\frac{x}{x+1}\right)^{-1} \quad \quad \quad\left[e^{\log x}=x\right]$

$=\frac{x+1}{x}$

So, the solution is given by

\begin{aligned} &=y(I f)=\int Q(I f) d x+C \\\\ &=y \times\left(\frac{x+1}{x}\right)=\int \frac{x+1}{x} d x+C \\\\ &=\left(\frac{x+1}{x}\right) y=\int \frac{x}{x} d x+\int \frac{1}{x} d x+C \end{aligned}

\begin{aligned} &=\left(\frac{x+1}{x}\right) y=\int 1 d x+\int \frac{1}{x} d x+C \\\\ &=\left(\frac{x+1}{x}\right) y=x+\log x+C \ldots(i) \end{aligned}

Since curve passes through the point $\left ( 1,0 \right )$ it satisfies the equation of the curve

$=\left(\frac{1+1}{1}\right) 0=1+\log 1+C$

$=0=1+0+C \quad[\log 1=0]$

$=C=-1$

Substituting value of C in equation (i) we get

\begin{aligned} &=\left(\frac{x+1}{x}\right) y=x+\log x-1 \\\\ &=y=\left(\frac{x}{x+1}\right)(x+\log x-1) \end{aligned}

Hence the required equation for the curve is found.