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  • Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 59

Need solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.7 question 59

Answers (1)

Answer: (1+\log x)^{2}=\log \left(1-y^{2}\right)^{2}+1

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

Given: \left(1-y^{2}\right)(1+\log x) d x+2 x y d y=0 \text { given that } y=0 \text { when } x=1

Solution:

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

          Integrating both sides

        \begin{aligned} &\int \frac{1+\log x}{x} d x=\int \frac{-2 y}{1-y^{2}} d y \ldots \ldots \ldots \ldots . .(*) \\ &\Rightarrow I_{1}=I_{2} \text { where } I_{1}=\int \frac{1}{x}(1+\log x) d x \text { and } \\\\ &I_{2}=\int \frac{-2 y}{1-y^{2}} d y \\\\ &\therefore I_{1}=\int \frac{1}{x}(1+\log x) d x \end{aligned}

        Put

        \begin{aligned} &1+\log x=t \\\\ &\frac{1}{x} d x=d t \\\\ &=\int t d t \\\\ &=\frac{t^{2}}{2}+c=\frac{(1+\log x)^{2}}{2}+c \end{aligned}

        Now,   I_{2}=\int \frac{-2 y}{1-y^{2}} d y

        Put

        \begin{aligned} &1-y^{2}=t \\\\ &-2 y d y=d t \\\\ &=\int \frac{d t}{t} \\\\ &=\log |t|+c \\\\ &=\log \left|1-y^{2}\right|+c \end{aligned}

        Put the values in (*)  we get

        \frac{(1+\log x)^{2}}{2}=\log \left|1-y^{2}\right|+c                ................(1)

        Now according to given y = 0 when x = 1

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

        \begin{aligned} &\Rightarrow 1=2[(0)+c] \\\\ &\Rightarrow 1=2 c \Rightarrow c=\frac{1}{2} \end{aligned}

        Put in (1) we get

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

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