Chapter 6.4, Problem 1E

To determine

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

Answer to Problem 1E

  $y=\sqrt{x^2+3}$ .

The solution is valid for all the real numbers.

Explanation of Solution

Given:

  $\frac{dy}{dx}=\frac{x}{y}$ and $y=2$ 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,

  $\frac{dy}{dx}=\frac{x}{y}$

By separating the variables, it follows that

  $ydy=xdx$

Now by integrating on both sides, it follows that

  $\begin{array}{c}{\displaystyle \int ydy}={\displaystyle \int xdx}\\ \frac{y^2}{2}=\frac{x^2}{2}+C_1\\ y^2=x^2+2C_1\\ y^2=x^2+C,\text{where}2C_1=C\end{array}$

Now by using the given initial conditions, it follows that

  $\begin{array}{c}{2}^2=1^2+C\\ 4=1+C\\ C=3\end{array}$

Now put $C=3$ in $y^2=x^2+C$ , it follows that

  $\begin{array}{c}{y}^2=x^2+3\\ y=\sqrt{x^2+3}\end{array}$

Since $x^2+3>0$ for any real value of $x$ , therefore the solution is valid for all the real numbers.