Question

In Problems 31-34 find an explicit solution of the given initial-value problem. Determine the exact intervalof definition by analytical methods. Use a graphing utility to plot the graph of the solution.

$\mathbf{sinxdx}\mathbf{+}\mathbf{ydy}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The explicit solution & exact interval are$\mathrm{y}=\sqrt{2\mathrm{cosx}-1}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{I}=\left(-\frac{\mathrm{\pi }}{3},\frac{\mathrm{\pi }}{3}\right)$ respectively.

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.

Integrate differential equation

Consider the differential equation$\sin xdx+ydy=0$

Separate the variables by subtracting the$dx$ term to the right hand side.

$ydy=-\sin xdx$

Now we integrate the left hand side in terms of $y$and the right hand side in terms of $x$.

$\begin{array}{c}\int ydy=\int -\sin xdx\\ \frac{1}{2}{y}^2=\cos x+c\end{array}$

Apply initial condition

Substitute the initial condition $y\left(0\right)=1$in order to solve for $c$.

$\begin{array}{c}2\text{'}{\left(1\right)}^2=\cos 0+c\\ \frac{1}{2}=1+c\\ -\frac{1}{2}=c\end{array}$

The solution becomes$\frac{1}{2}{y}^2=\cos x-\frac{1}{2}$

Find explicit solution

We need to solve for$y$ in order to get the explicit solution.

$\begin{array}{c}\frac{1}{2}{y}^2=\cos x-\frac{1}{2}\\ y^2=2\cos x-1\\ y=\pm \sqrt{2\cos x-1}\end{array}$

The initial condition is $\left(0,1\right)$. So,

$\begin{array}{c}y=\pm \sqrt{2\cos \left(0\right)-1}\\ =\pm \sqrt{2-1}\\ =\pm \sqrt{1}\\ =\pm 1\end{array}$

Since$y=1$ we choose the positive square root and have the explicit solution,

$y=\sqrt{2\cos x-1}$

Graph

Find interval

The solution is only defined when the term inside the square root is defined.

$\begin{array}{c}2\mathrm{cosx}-1\ge 0\\ 2\mathrm{cosx}\ge 1\\ \mathrm{cosx}\ge \frac{1}{2}\end{array}$

Recall that the reference angle for$\cos x=1/2$ is $x=\pi /3$. So $\cos x\ge 1/2$for intervals$-\frac{\pi }{3}+2\pi n\le x\le \frac{\pi }{3}+2\pi n$ for all integers $n$.

Since the initial condition occurs atrole="math" localid="1667810537069" $x=f\left(0\right)$

The graph of the solution we are interested in is the domain $I=\left(-\frac{\pi }{3},\frac{\pi }{3}\right)$.