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

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

Answers (1)

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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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