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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 14

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 14

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Answer: 2 x y-2 x-y-2=0

Given: The differential equation is y-x \frac{d y}{d x}=y^{2}+\frac{d y}{d x} .

To find: We have to find the equation of the curve which passes through \left ( 2,2 \right )

Hint: We will integrate the given differential equation and then we use the points \left ( 2,2 \right )

Solution: we have,

        =y-x \frac{d y}{d x}=y^{2}+\frac{d y}{d x}

Integrating on both sides we get,

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

        \begin{aligned} &=>\int \frac{d y}{y(1-y)}=\int \frac{d x}{1+x} \\\\ &=>\int\left[\frac{1}{y}-\frac{1}{1-y}\right] d y=\int \frac{d x}{x+1} \end{aligned}

        =>\log |y|-\log |1-y|=\log |x+1|+C \ldots(i)

Since the curve passing through the point \left ( 2,2 \right ) it satisfies equation (i)

Then,

        \begin{aligned} &\log |2|-\log |1-2|=\log |2+1|+C \\\\ &\log |2|-\log |-1|=\log |3|+C \\\\ &C=\log |2|-\log |3| \\\\ &C=\log \left|-\frac{2}{3}\right| \end{aligned}

Putting the value of C in equation (i) we get

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

Taking antilogarithm on both sides,

        \begin{aligned} &\frac{y}{1-y}=-\frac{2(x+1)}{3} \\\\ &3 y=-2(x+1)(1-y) \\\\ &3 y=-2(1+x-y-x y) \\\\ &3 y=-2-2 x+2 y+2 x y \end{aligned}

        \begin{aligned} &3 y+2+2 x-2 y-2 x y=0 \\\\ &y+2+2 x-2 x y=0 \\\\ &2 x y-2 x-2-y=0 \end{aligned}

Hence, required equation is found.

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