Chapter 6.4, Problem 9E

To determine

To solve: the given initial value problem by using the separation of variables.

Answer to Problem 9E

  $y=\frac{1}{x^2+3}$ .

The solution is valid for all the real numbers.

Explanation of Solution

Given:

  $\frac{dy}{dx}=-2x{y}^2$ and $y=0.25$ when $x=1$ .

Concept used:

A differential equation of the form $\frac{dy}{dx}=f\left(y\right)g\left(x\right)$ is called separable.

The variables can be separated by writing it in the form $\frac{1}{f\left(y\right)}dy=g\left(x\right)dx$ .

The solution is found by anti differentiating each side with respect to its thusly isolated variable.

Calculation:

Consider the given differential equation,

  $\frac{dy}{dx}=-2x{y}^2$

By separating the variables, it follows that

  $\begin{array}{c}\frac{dy}{y^2}=-2xdx\\ y^{-2}dy=-2xdx\end{array}$

Now by integrating on both sides, it follows that

  $\begin{array}{c}{\displaystyle \int y^{-2}dy}={\displaystyle \int -2xdx}\\ -\frac{1}{y}=-x^2+C\end{array}$

Now by using the given initial conditions, it follows that

  $\begin{array}{c}-\frac{1}{0.25}=-{\left(1\right)}^2+C\\ -4=-1+C\\ C=-3\end{array}$

Now put $C=-3$ in $-\frac{1}{y}=-x^2+C$ , it follows that

  $\begin{array}{c}-\frac{1}{y}=-x^2-3\\ \frac{1}{y}=x^2+3\\ y=\frac{1}{x^2+3}\end{array}$

Clearly, the solution is valid for all the real numbers.