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Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 21 maths textbook solution

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Answer: -\log |y|+\log |y-1|=-\frac{1}{2} \log \left|1-x^{2}\right|+\log c

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

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

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

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

          Integrating both sides

        \int \frac{d y}{y(y-1)}=\int \frac{x d x}{\left(1-x^{2}\right)}

        \int\left(\frac{1}{y-1}-\frac{1}{y}\right) d y=\frac{-1}{2} \int \frac{-2 x}{\left(1-x^{2}\right)} d x


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

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

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