Question

In Problems 25–28 use (12) to verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval I of definition of each solution.

$x\frac{dy}{dx}-3xy=1;y=e^{3x}{\int }_1^x\frac{e^{-3t}}{t}dt$

Short answer

The indicated function is a solution of the differential function.

Step-by-step solution

Simplify the given differential equation.

Let the given differential equation be $y=e^{3x}{\int }_1^x\frac{e^{-3t}}{t}dt$.

Multiply each side of the equation by $e^{-3x}$.

$$\begin{gathered} y{e}^{-3x}=e^{3x}{e}^{-3x}{\int }_1^x\frac{e^{-3t}}{t}dt \\ y{e}^{-3x}={\int }_1^x\frac{e^{-3t}}{t}dt \end{gathered}$$

Determine the solution of the indicated function.

Take differential on both sides of the equation.

$$\begin{gathered} \frac{d}{dx}\left(y{e}^{-3x}\right)=\frac{d}{dx}{\int }_1^x\frac{e^{-3t}}{t}dt\left(\because F\text{'}\left(x\right)=\frac{d}{dx}{\int }_a^{x}g\left(t\right)=g\left(x\right)\right) \\ {e}^{-3x}\frac{dy}{dx}-3y{e}^{-3x}=\frac{e^{-3x}}{x} \end{gathered}$$

Multiply$x$ on both sides of the equation.

$x{e}^{-3x}\frac{dy}{dx}-3yx{e}^{-3x}=e^{-3x}$

Divide$e^{-3x}$ on both sides of the equation.

$x\frac{dy}{dx}-3yx=1$

Hence, the indicated function is a solution of the differential function.