Question

Answer Problems 1-12without referring back to the text. Fill in the blanks or answer true or false.

$\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{x}}\stackrel{\mathbf{\text{'}}}{{\mathbf{O}}_{\mathbf{0}}}\mathit{t}{\mathit{e}}^{\mathbf{-}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{}\mathbf{t}}\mathit{d}\mathit{t}$ is a solution of the linear first-order differential equation.

Short answer

The differential equation is $\frac{dy}{dx}=x-y\sin \left(x\right)$.

Step-by-step solution

Definition

The standard form of a linear differential equation is $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}\mathbf{-}\mathit{P}\mathit{y}\mathbf{+}\mathit{Q}$, and it contains the variabley, and its derivatives.

Simplify equation

We are given the solution to a differential equation and asked to find the differential equation. In order to solve this we can think of undoing the strategy of integrating factor. We start by isolating the integral on the right hand side.

$$\begin{gathered} y=e^{\cos x}{\int }_0^{x}t{e}^{-\cos t}dt \\ y{e}^{\cos x}={\int }_0^{x}t{e}^{-\cos t}dt \end{gathered}$$

Solve equation

The definite integral on the right hand side is equivalent to the indefinite integral of the same function.

$\begin{array}{rcl}y{e}^{-\cos x}& =& \int x{e}^{-\cos x}dx\\ \frac{d}{dx}\left(y{e}^{-\cos x}\right)& =& x{e}^{-\cos x}\\ \frac{dy}{dx}{e}^{-\cos x}+y\sin \left(x\right)e^{-\cos x}& =& x{e}^{-\cos x}\end{array}$

Now we can see that the integrating factor is e^{-cos x}, which we can divide out of the expression.

$$\begin{gathered} \frac{dy}{dx}+y\sin \left(x\right)=x \\ \frac{dy}{dx}=x-y\sin \left(x\right) \end{gathered}$$