Question

Reread Example 4 and find the general solution of the differential equation on the interval (-3, 3)

Short answer

$y=\frac{c}{\sqrt{9-x^2}}$

Step-by-step solution

Definition of initial-value problem and differential equation

The unknown function $y\left(x\right)$and its derivatives at a number$x_0$ . On some interval I containing$x_0$ the problem of solving an nth-order differential equation subject to n side conditions specified at:

Solve:$\frac{d^{n}y}{d{x}^n}=f(x,y,y\text{'},...,y^{(n-1)})$

Subject to:$y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,y^{(n-1)}\left(x_0\right)=y_{(n-1)}.$

Where $y_0,y_1,...,y_{n-1}$ are arbitrary constants, is called n-th order Initial Value Problem (IVP). The values of $y\left(x\right)$and its first n-1 derivatives at$x_0$$y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,y^{(n-1)}\left(x_0\right)=y_{(n-1)}.$are called Initial Conditions.

Differential Equation is defined as the equation that consists of the derivatives of one or more dependent functions in respect to one or more independent functions.$dy/dx=f\left(x\right)$

Determination of differential equation on (-3, 3)

The given interval (-3, 3)

The integrating factor will be $e^{\int P\left(x\right)dx}$

Solve the equation using the interval (-3,3)

$$\begin{gathered} {}^{\frac{x}{x^2-9}dx}=e^{\frac{1}{2}\int \frac{2x}{x^2-9}dx} \\ =e^{\frac{1}{2}\ln \left|x^2-9\right|} \\ =e^{\frac{1}{2}\ln \left(9-x^2\right)} \\ =\sqrt{9-x^2} \end{gathered}$$

Multiplying d/dx in the above function (d/dx=standard form)

$\frac{d}{dx}\left[\sqrt{9-x^2}.y\right]=0$

Integrating both sides;

$\sqrt{9-x^2}.y=c$

The general solution on interval (-3, 3) is; $y=\frac{c}{\sqrt{9-x^2}}$