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

Answers (1)


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


        \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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