Question

In Problems 3–6,$\mathit{y}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathit{c}\mathbf{)}$ is a one-parameter family of solutions of the first-order DE$\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{x}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{0}$. Find a solution of the first-order IVP consisting of this differential equation and the given initial condition. Give the largest interval I over which the solution is defined.

$\mathit{y}\left(0\right)\mathbf{=}\mathbf{1}$

Short answer

The solution of the differential equation is $y=\frac{1}{x^2+1}\text{on}I:(-\infty ,\infty )$.

Step-by-step solution

Definition of Initial Value Problems

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-thorder 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.

Compute the value of c

Given $y=\frac{1}{x^2+c}$

Substitute the initial condition

$y\left(0\right)=1$

$$\begin{gathered} 1=\frac{1}{0^2+c}=\frac{1}{c} \\ c=1 \end{gathered}$$

Substitute the value of c

Substitute the value

$y=\frac{1}{x^2+1}$

The domain of the solution

$D:(-\infty ,\infty )$

Since$x^2$will never be negative, the denominator of the solution will never be equat to 0.

Therefore, the domain of the solution is all real numbers.

Therefore, the largest interval is$I:(-\infty ,\infty )$

Hence, thesolution of the differential equation is$\mathit{y}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}\mathbf{}\mathbf{}\mathbf{\text{on}}\mathbf{}\mathbf{}\mathit{I}\mathbf{:}\mathbf{(}\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{\infty }\mathbf{)}$.