Question

In problems 23-28 Find an explicit solution to the given initial-value problem.

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{{\text{y}}^{\text{2}}\text{- 1}}{{\text{x}}^{\text{2}}\text{- 1}}\text{,}\text{y(2) = 2}$

Short answer

$\frac{y-1}{y+1}=\frac{x-1}{x+1}$

Step-by-step solution

Definition of separable equation

A first-order differential equation of the form$\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is said to be separable or to have separable variables.

Separate the variables and integration

$\frac{dy}{dx}=\frac{y^2-1}{x^2-1}$

Separate the differentials.

$\frac{dy}{y^2-1}=\frac{dx}{x^2-1}$

Integrate both sides.

$\int \frac{dy}{y^2-1}=\int \frac{dx}{x^2-1}$

Use the integration Rule

$$\begin{gathered} \int \frac{du}{u^2-1}=\frac{1}{2}\ln \left|\frac{u-1}{u+1}\right|+C \\ \frac{1}{2}\ln \left|\frac{y-1}{y+1}\right|=\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+C \end{gathered}$$

Substitute the initial condition

Apply the initial condition

$y\left(2\right)=2\Rightarrow x=2;y=2$

Therefore

$$\begin{gathered} \frac{1}{2}\ln \left|\frac{2-1}{2+1}\right|=\frac{1}{2}\ln \left|\frac{2-1}{2+1}\right|+C \\ C=0 \end{gathered}$$

Substitute C You get

$$\begin{gathered} \frac{1}{2}\ln \left|\frac{y-1}{y+1}\right|=\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+0 \\ \ln \left|\frac{y-1}{y+1}\right|=\ln \left|\frac{x-1}{x+1}\right| \end{gathered}$$

Take e to both sides and simplify

$$\begin{gathered} {}^{\ln \left|\frac{y-1}{y+1}\right|}=e^{\ln \left|\frac{x-1}{x+1}\right|} \\ \frac{y-1}{y+1}=\frac{x-1}{x+1} \end{gathered}$$

Hence the solution is$\frac{\text{y - 1}}{\text{y + 1}}\text{=}\frac{\text{x - 1}}{\text{x + 1}}$