Question

Express the solution of the given initial-value

problem in terms of an integral-defined function.

$\mathbf{2}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{\left(}\mathbf{4}\mathbf{cosx}\mathbf{\right)}\mathbf{y}\mathbf{=}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$ dr=0

Short answer

The integral defined function of the given differential equation is

$y\left(x\right)=e^{-2\sin x}\left[1+\frac{1}{2}{\int }_0^{x}t{e}^{2\sin t}dt\right]$

Step-by-step solution

Covert the given differential equation into linear differential equation

The given differential equation, can be written as

$\frac{\mathrm{d}y}{\mathrm{d}x}+\left(2\cos x\right)y=\frac{x}{2}.....\left(1\right)$

Linear differential equation of the first order

${\mathrm{y}}^{\text{'}}+\mathrm{P}\left(\mathrm{x}\right)\mathrm{y}=\mathrm{f}\left(\mathrm{x}\right)...\left(2\right)$

Based on (1) and (2), we obtain

$\mathrm{P}\left(\mathrm{x}\right)=2\mathrm{cosx}\mathrm{and}\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{2}$

Find integrating factor

The integrating factor is given by

$e^{\int P\left(x\right)\text{d}x}\mathrm{...}\left(3\right)$

Use the formula (3), to get integrating factor

$e^{\int P\left(x\right)\text{d}x}=e^{2\int \cos x\text{d}x}=e^{2\sin x}$

Finding general solution

Thus, the solution is

$\begin{array}{c}{\mathrm{ye}}^{2\mathrm{sinx}}={\int }_0^{\mathrm{x}}{\mathrm{e}}^{2\mathrm{sint}}\frac{\mathrm{t}}{2}\text{d}\mathrm{t}+\mathrm{c}\\ \mathrm{y}={\mathrm{e}}^{-2\mathrm{sinx}}\frac{1}{2}{\int }_0^{\mathrm{x}}{\mathrm{e}}^{2\mathrm{sint}}\mathrm{t}\text{d}\mathrm{t}+{\mathrm{ce}}^{-2\mathrm{sinx}}\end{array}$

Find the value of c

Use the initial condition y(0) =1 to get c.

$\begin{array}{c}y\left(0\right)=\frac{1}{2}{e}^{-2\sin 0}\underset{x\to 0}{\lim }{\int }_0^x{e}^{2\sin t}t\text{d}t+c{e}^{-2\sin 0}(\sin 0=0,e^0=1)\\ 1=0+c\\ c=1\end{array}$

Substitute the value $\mathit{C}$ in the equation

$y\left(x\right)=e^{-2\sin x}\left[1+\frac{1}{2}{\int }_0^{x}t{e}^{2\sin t}\text{d}t\right]$