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

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)

        =\frac{(-4)^{2}}{2}=-\frac{3^{2}}{2}+C

        \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

        =>\frac{y^{2}}{2}=-\frac{x^{2}}{2}+\frac{25}{2}

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

        =x^{2}+y^{2}=25

Hence, found.

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