Solve the initial-value problem dy/dx=-x/y, y(4)=-3.

(1) $\begin{align*} \frac{dy}{dx} = -\frac{x}{y}, y(4) = -3 \end{align*}$

Now

$\begin{align*} ydy &= -xdx\\ \int ydy &= -\int xdx\\ y^2 \times \frac{1}{2} &= -x^2 \times \frac{1}{2} + C_1 \end{align*}$

$\begin{align*} \frac{1}{2}(-3)^2 &= -\frac{1}{2}(4)^2 + C_1 \end{align*}$

$\begin{align*}\frac{9}{2} &= -8 + C_1, \end{align*}$

$\begin{align*}\ C_1 = \frac{25}{2} \end{align*}$

$\begin{align*} \frac{1}{2} y^2 &= -\frac{1}{2} x^2 + \frac{25}{2} \end{align*}$

$\begin{align*} y^2 &= -x^2 + 25 \end{align*}$

$\begin{align*}y &= \pm\sqrt{25-x^2} \end{align*}$

Using the initial condition $y(4)=-3$ , we get:

$\begin{align*} y &= -\sqrt{25-4^2} \end{align*}$

$\begin{align*} y &= -\sqrt{9} \end{align*}$

$\begin{align*}y &= -3 \end{align*}$

Therefore, the solution to the initial value problem is:

$\begin{equation*} y = -\sqrt{25-x^2} \end{equation*}$