Question

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

$\mathit{y}\text{'}\mathbf{-}\mathbf{\left(}\mathbf{sinx}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{2}\mathbf{sinx}\mathbf{,}\mathbf{y}\left(\frac{\mathbf{\pi }}{\mathbf{2}}\right)\mathbf{=}\mathbf{1}$

Short answer

So, the largest intervalon which the solution is defined is$(-\infty ,\infty )$.

. And the solution is $y=-2+c{e}^{-\cos x}$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

$\frac{dy}{dx}-\left(\sin x\right)y=2\sin x,y\left(\frac{\pi }{2}\right)=1\dots \dots \left(1\right)$

This is a linear differential equation of first order

Compare it with$\frac{dy}{dx}+p\left(x\right)y=q\left(x\right)$

Observe that, $p\left(x\right)=-\sin x$and$q\left(x\right)=2\sin x$

To solve (1), it is required to multiply the differential equation with an integrating factor.

Integrating factor$=\mu \left(x\right)=e^{\int m\left(x\right)dx}$

$$\begin{gathered} =e^{\int -\sin udx} \\ =e^{\cos x} \end{gathered}$$

Multiply,$e^{\cos x}\left(\frac{dy}{dx}-\left(\sin x\right)y\right)=e^{\cos x}\left(2\sin x\right)$

It automatically get simplified as$d\left(y{e}^{\cos x}\right)=2\sin x·e^{\cos x}\dots \dots \left(2\right)$

Integrate both sides

$e^{\cos x}y=\int 2\sin x{e}^{\cos x}dx+c$

In view of the change of variable, suppose$\cos x=U$

$-Sinx\text{d}x=\text{d}u$

Use these in the above integral,

role="math" localid="1668496209464" $$\begin{gathered} e^{\cos x}y=-2\int e^{x}du+c \\ {e}^{cosx}y=-2e^{\cos x}+c \\ or,y=-2+c{e}^{-\cos x}\dots \dots \left(3\right) \end{gathered}$$

This is the general solution.

Substitute initial condition

To find the particular solution, substitute the initial condition$y\left(\frac{\pi }{2}\right)=1$in the general solution.

That is$\text{I}=-2+c{e}^{-\infty -\frac{\pi }{2}}$

Solve it to get c =3

Substitute this in the general solution, the required solution of the initial value problem is$y=-2+3e^{-cosx}$.

The solution is defined for all $x\in (-\infty ,\infty )$.