Question

Solve the given initial-value problem. Give the largest interval / over which the solution is defined.

$\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{\left(}\mathbf{tanx}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{co}{\mathbf{s}}^{\mathbf{2}}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$

Short answer

So, the largest intervalon which the solution is defined is$-\frac{\pi }{2}<x<\frac{\pi }{2}$.

. And the solution is $y\left(x\right)=\sin x\cos x-\cos x$for the given initial problem.

Step-by-step solution

Form of first order differential equation

A first order linear differential equation is a differential equation of the form $y\text{'}+p\left(x\right)y=q\left(x\right)y\text{'}+p\left(x\right)y=q\left(x\right)y\text{'}+p\left(x\right)y=q\left(x\right)$.

Evaluation

Consider the initial value problem,

$y^{\text{'}}+\left(\tan x\right)y={\cos }^{2}x,y\left(0\right)=-1$

The objective is to solve the initial value problem and then give a largest intervalover which the solution is defined.

By comparing the differential equation with the standard form,

$$\begin{gathered} y^{\text{'}}+P\left(x\right)y=Q\left(x\right) \\ P\left(x\right)=\tan x,Q\left(x\right)={\cos }^{2}x \end{gathered}$$

Note that the functions $P\left(x\right)=\tan x,Q\left(x\right)={\cos }^{2}x$are continuous on $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

Find the Integrated factor

Observe that, given initial value problem is a first order linear differential equation.

Integrating factor of it is,

$$\begin{gathered} .\text{F}=e^{\int \tan xdx} \\ =e^{\ln \left|\sec x\right|} \\ =\left|\sec x\right| \\ \text{since}\sec x>0\text{for}x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right) \\ \text{so}\left|\sec x\right|\text{can be rereplaced with sec}x \\ =\sec x \end{gathered}$$

Multiply the given differential equation with the integrating factor,

$$\begin{gathered} \sec x\left[y^{\text{'}}+\left(\tan x\right)y\right]=\sec x·{\cos }^{2}x \\ y\sec x=\int \cos xdx \\ y\sec x=\sin x+c \\ y=\sin x\cos x+c\cos x\text{for}-\frac{\pi }{2}<x<\frac{\pi }{2} \end{gathered}$$

Find the value of constant

To obtain the value of integration constant use the initial condition

$$\begin{gathered} y\left(0\right)=\sin \left(0\right)\cos \left(0\right)+c\cos \left(0\right) \\ -1=0·1+c·1 \\ -1=c \end{gathered}$$

Hence, the required solution is $y\left(x\right)=\sin x\cos x-\cos xfor-\frac{\pi }{2}<x<\frac{\pi }{2}$ .