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

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.