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Please solve RD Sharma class 12 chapter Differential Equations exercise 21.7 question 9 maths textbook solution

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Answer: y-1=C x y

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

Given: x \frac{d y}{d x}+y=y^{2}

Solution: x \frac{d y}{d x}+y=y^{2}

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

         Integrating on both sides       

        \int \frac{d y}{y(y-1)}=\int \frac{d x}{x} \\\\

        \begin{aligned} &\int \frac{y-(y-1)}{y(y-1)} d y=\int \frac{d x}{x} \\\\ &\int \frac{1}{(y-1)} d y-\int \frac{1}{y} d y=\int \frac{d x}{x} \end{aligned}

        We get,

        \begin{aligned} &-\log y+\log (y-1)=\log x+\log c \\\\ &\log \frac{(y-1)}{y}=\log x C \\\\ &\frac{y-1}{y}=C x \\\\ &y-1=C x y \end{aligned}

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