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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

Answers (1)

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.

Posted by

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