Question

In Problems 29 and 30 proceed as in Example 5 and find an explicit solution of the given initial value problem.

$\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{=}{\mathit{y}}^{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{y}\mathbf{(}\mathbf{-}\mathbf{2}\mathbf{)}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}$

Short answer

The explicit solution is $y\left(x\right)=\frac{1}{3-{\int }_{-2}^x\sin t^{2}dt}$.

Step-by-step solution

Definition

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

Separate differential equation

Consider the initial value problem, $\frac{dy}{dx}=y^2\sin x^{2}andy(-2)=\frac{1}{3}$.

Rewrite the equation as,

$$\begin{gathered} \frac{dy}{dx}=y^2\sin x^2 \\ \frac{1}{y^2}\frac{dy}{dx}=\sin x^2 \\ Letg\left(x\right)=\sin x^2 \end{gathered}$$

The function $g\left(x\right)=\sin x^2$is continuous on$(-\infty ,\infty )$, but its anti derivative is not an elementary function.

Use t is a dummy variable of integration.

Separate the variables to differential equation.

$\begin{array}{rcl}\frac{dy}{dx}& =& y^2\sin x^2\\ \frac{1}{y^2}\frac{dy}{dx}& =& \sin x^2\end{array}$

And the initial condition is$y(-2)=\frac{1}{3}$.

Solve integral

Integrate both sides with respect tousing the limits $[-2,x]$.

$\begin{array}{rcl}{\int }_{-2}^x\frac{1}{y^2}\frac{dy}{dt}dt& =& {\int }_{-2}^x\sin t^{2}dt\\ {\left.-\frac{1}{y\left(t\right)}\right]}_{-2}^x& =& {\int }_{-2}^x\sin t^{2}dt\\ -\frac{1}{y\left(x\right)}+\frac{1}{y(-2)}& =& {\int }_{-2}^x\sin t^{2}dt\\ -\frac{1}{y\left(x\right)}+\frac{1}{\frac{1}{3}}& =& {\int }_{-2}^x\sin t^{2}dt\end{array}$

Apply initial condition

Substitute$y(-2)=\frac{1}{3}$

$\begin{array}{rcl}-\frac{1}{y\left(x\right)}+3& =& {\int }_{-2}^x\sin t^{2}dt\\ -\frac{1}{y\left(x\right)}& =& -3+{\int }_{-2}^x\sin t^{2}dt\\ \frac{1}{y\left(x\right)}& =& 3-{\int }_{-2}^x\sin t^{2}dt\\ y\left(x\right)& =& \frac{1}{3-{\int }_{-2}^x\sin t^{2}dt}\end{array}$

Therefore, the explicit solution of the given initial value problem is $y\left(x\right)=\frac{1}{3-{\int }_{-2}^x\sin t^{2}dt}$.