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Please solve RD Sharma class 12 chapter Differential Equations exercise 21.11 question 13 maths textbook solution

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Answer: x^2+y^2=25

Given: The slope of tangent at a point  P(x, y)=-\frac{x}{y}

To find: We have to find the equation of the curve which passes through \left ( 3,-4 \right )

Hint: First take the slope of the curve and integrate the equation.

Solution: So,\frac{d y}{d x}=-\frac{x}{y}

              =y d y=-x d x

Integrating on both sides we get,

        \begin{aligned} &=\int y d y=-\int x d x \\\\ &=\frac{y^{2}}{2}=-\frac{x^{2}}{2}+C \ldots(i) \end{aligned}

Where C is constant

Since the curve passes through the point \left ( 3,-4 \right ) such that \left ( 3,-4 \right ) satisfy equation (i)


        \begin{aligned} &=>\frac{16}{2}=-\frac{9}{2}+C \\\\ &=>8=-\frac{9}{2}+C \\\\ &=>C=8+\frac{9}{2} \end{aligned}

        \begin{aligned} &=>C=\frac{16+9}{2} \\\\ &=>C=\frac{25}{2} \end{aligned}

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


        =y^{2}=-x^{2}+25        [Multiplying by 2 on both sides]


Hence, found.

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