Chapter 6.3, Problem 29E

To determine

To find: The value of differential equation.

Answer to Problem 29E

The answer is:

  $e^{4x}\left(\frac{1}{4}{x}^2-\frac{1}{8}x+\frac{1}{32}\right)+C$

Explanation of Solution

Given information:

The differential equation is: $\frac{dy}{dx}=x^2{e}^{4x}$

Calculation:

The given differential equation is: $\frac{dy}{dx}=x^2{e}^{4x}$

Put the variables in a separated row with the $y$ s on the left and the $x$ s on the right. Integrate the left side after writing an integral symbol on both sides:

  $\begin{array}{c}\frac{dy}{dx}=x^2{e}^{4x}\\ dy=x^2{e}^{4x}dx\\ \int dy=\int x^2{e}^{4x}dx\\ y=\int x^2{e}^{4x}dx\end{array}$

Use tabular integration with

  $\begin{array}{c}f(x)=x^2\\ g(x)=e^{4x}t\end{array}$

to integrate the right side.

Find the anti derivatives of $e^{4x}$ and the derivatives of $x^2$ till eventually reach 0. Then, draw arrows connecting each first-column expression (except from the final expression of 0) to the expression in the second-column row below. Start with a positive sign and alternate between positive and negative signs as write the signs above the arrows.

  

Utilize the operation signals located above each arrow to combine the results of the functions that are connected by arrows. Remember to add the $+C$ :

  $\begin{array}{c}\int x^2{e}^{4x}dx=x^2\left(\frac{1}{4}{e}^{4x}\right)-2x\left(\frac{1}{16}{e}^{4x}\right)+2\left(\frac{1}{64}{e}^{4x}\right)+C\\ =\frac{1}{4}{x}^2{e}^{4x}-\frac{1}{8}x{e}^{4x}+\frac{1}{32}{e}^{4x}+C\\ =e^{4x}\left(\frac{1}{4}{x}^2-\frac{1}{8}x+\frac{1}{32}\right)+C\end{array}$

Therefore the required value of the given differential equation $\frac{dy}{dx}=x^2{e}^{4x}$ is

  $e^{4x}\left(\frac{1}{4}{x}^2-\frac{1}{8}x+\frac{1}{32}\right)+C$