#### Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 53 maths textbook solution

Answer: $y=x+\log \left(\frac{x y+2 x}{3}\right)^{2}+2$

Hint: Separate the terms of x and y and then integrate them.

Given: $x y \frac{d y}{d x}=(x+2)(y+2) ; p t(1,-1)$

Solution:

\begin{aligned} &x y \frac{d y}{d x}=(x+2)(y+2) \\\\ &\Rightarrow \frac{y}{y+2} d y=\frac{x+2}{x} d x \end{aligned}

Integrating both sides

\begin{aligned} &\int \frac{y}{y+2} d y=\int \frac{x+2}{x} d x \\\\ &\int \frac{y+2-2}{y+2} d y=\int\left(1+\frac{2}{x}\right) d x \\\\ &\Rightarrow \int\left(1-\frac{2}{y+2}\right) d y=\int\left(1+\frac{2}{x}\right) d x \end{aligned}

\begin{aligned} &\Rightarrow y-2 \log |y+2|=x+2 \log |x|+c \\\\ &\Rightarrow y=x+2 \log |x|+2 \log |y+2|+c \\\\ &\Rightarrow y=x+2 \log |x(y+2)|+c \\\\ &\Rightarrow y=x+\log (x y+2 x)^{2}+c \end{aligned}        ..............(1)

It passes through pt(-1,1)

\begin{aligned} &1=-1+\log |(-1)(1)+2(-1)|^{2}+c \\\\ &\Rightarrow 2=\log (-1-2)^{2}+c \\\\ &\Rightarrow 2-\log 9=c \end{aligned}

Put in (1) we get

\begin{aligned} &\Rightarrow y=x+\log (x y+2 x)^{2}+2-\log 9 \\\\ &\Rightarrow y=x+\log \left(\frac{x y+2 x}{3}\right)^{2}+2 \\\\ &{\left[\therefore \log m-\log n=\log \frac{m}{n}\right]} \end{aligned}

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