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  • Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 45 sub question (ix) maths textbook solution

Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 45 sub question (ix) maths textbook solution

Answers (1)

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

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

Given: 2(y+3)-x y \frac{d y}{d x}=0, y(1)=-2

Solution:

        \begin{aligned} &2(y+3)-x y \frac{d y}{d x}=0 \\\\ &\Rightarrow x y \frac{d y}{d x}=2(y+3) \Rightarrow \frac{y d y}{y+3}=\frac{2}{x} d x \end{aligned}

        Integrating both sides

        \begin{aligned} &\int \frac{y d y}{y+3}=\int \frac{2}{x} d x \\\\ &\Rightarrow \int \frac{y+3-3}{y+3} d y=2 \int \frac{1}{x} d x \\\\ &\Rightarrow \int\left(1-\frac{3}{y+3}\right) d y=2 \int \frac{1}{x} d x \end{aligned}

        \begin{aligned} &\Rightarrow y-3 \log |y+3|=2 \log |x|+c \\\\ &\Rightarrow y=3 \log |y+3|+2 \log |x|+c \\\\ &\Rightarrow y=\log \left|(y+3)^{3}\right|+\log x^{2}+c \\\\ &\Rightarrow y=\log \left|(y+3)^{3} x^{2}\right|+c \end{aligned}        ..............(1)

        Given that y(1)=-2 \text { i.e. at } x=1, y=2

        \begin{aligned} &\therefore-2=\log \left|(-2+3)^{3}(1)^{2}\right|+c \\\\ &-2=\log 1+c \Rightarrow-2=0+c \Rightarrow c=-2 \\\\ &{[\therefore \log 1=0]} \end{aligned}

        Put in (1) we get

        y=\log \left|(y+3)^{3} x^{2}\right|-2

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