Question

In Exercises \(1-10\) , use separation of variables to solve the initial value problem. Indicate the domain over which the solution is valid. \(\frac{d y}{d x}=-\frac{x}{y} \quad\) and \(y=3\) when \(x=4\)

Short answer

The solution of the initial value problem is given by \(y^{2} = -x^{2} + 10\), which is valid for \(-\sqrt{10} \leq x \leq \sqrt{10}\)

Step-by-step solution

Separate Variables

For the given differential equation, \(\frac{d y}{d x}=-\frac{x}{y}\) the variables can be separated by cross multiplying to give \(y \, dy = - x \, dx\)

Integrate

Now, integrate both sides of the equation. This gives \(\int y \, dy = -\int x \, dx\), which can be integrated to yield \(\frac{y^{2}}{2} = -\frac{x^{2}}{2} + C\)

Solve for C

Replace the initial values \(y=3\) when \(x=4\) into the equation: \(\frac{3^{2}}{2} = -\frac{4^{2}}{2} + C\). Solve for \(C\) and you find that \(C = 5\)

Write out the general solution

Inserting \(C\) back into the equation yields the general solution: \(\frac{y^{2}}{2} = -\frac{x^{2}}{2} + 5\) or \(y^{2} = -x^{2} + 10\)

Determine the domain

Since \(y\) is a real number, \(y^{2}\) should be a non-negative number, which implies that \(-x^{2} + 10 \geq 0\). Solve this inequality to find the domain: \(-x^{2} \geq -10\), yielding that \(x^{2} \leq 10\) or \(-\sqrt{10} \leq x \leq \sqrt{10}\)