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Found in: Page 442

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Solve the differential equation $\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}{\mathbf{+}}{\mathbf{3}}{\mathbit{x}}{\mathbf{=}}{\mathbf{7}}$ and find the solution of the differential equation.

The solution is $f\left(t\right)=\frac{7}{3}+\frac{7}{3}{e}^{-3t}C$ .

See the step by step solution

## Step1: Definition of first order linear differential equation

Consider the differential equation ${f}{\text{'}}\left(t\right){-}{a}{f}\left(t\right){=}{g}\left(t\right)$ where ${g}\left(t\right)$ is a smooth function and ${\text{'}}{a}{\text{'}}$ is a constant. Then the general solution will be ${f}\left(t\right){=}{{e}}^{at}\int {e}^{-at}g\left(t\right)dt$.

## Step2: Determination of the solution

Consider the differential equation as follows.

$x\text{'}\left(t\right)+3x\left(t\right)=7$

Now, the differential equation is in the form as follows.

$f\text{'}\left(t\right)-af\left(t\right)=g\left(t\right)$, where $g\left(t\right)$ is a smooth function, then the general solution will be as follows.

$f\left(t\right)={e}^{at}\int {e}^{-at}g\left(t\right)dt$

## Step3: Compute the calculation of the solution

Substitute the value$7$ for $g\left(t\right)$ and $-3$ for in $f\left(t\right)={e}^{at}\int {e}^{-at}g\left(t\right)dt$ as follows.

$\begin{array}{l}f\left(t\right)={e}^{at}\int {e}^{-at}g\left(t\right)dtA\\ f\left(t\right)={e}^{-3t}\int {e}^{3t}×7×dt\\ f\left(t\right)=7{e}^{-3t}\int {e}^{3t}dt\end{array}$

Using substitution method in the integral as follows.

$\begin{array}{c}y=3t\\ dy=3dt\\ \frac{dy}{3}=dt\end{array}$

Substitute the value $3t$ for$y$ and $\frac{dy}{3}$ for $dt$ in $f\left(t\right)=7{e}^{3t}\int {e}^{-3t}dt$ as follows.

$\begin{array}{l}f\left(t\right)=7{e}^{-3t}\int {e}^{3t}dt\\ f\left(t\right)=7{e}^{-3t}\int {e}^{y}\frac{dy}{3}\\ f\left(t\right)=\frac{7}{3}{e}^{-3t}\left({e}^{y}+C\right)\end{array}$

Now, undo the substitution as follows.

$\begin{array}{c}f\left(t\right)=\frac{7}{3}{e}^{-3t}\left({e}^{y}+C\right)\\ f\left(t\right)=\frac{7}{3}{e}^{-3t}\left({e}^{3t}+C\right)\\ =\frac{7}{3}{e}^{-3t}{e}^{3t}+\frac{7}{3}{e}^{-3t}C\\ f\left(t\right)=\frac{7}{3}+\frac{7}{3}{e}^{-3t}C\end{array}$

Hence, the solution for the linear differential equation$x\text{'}\left(t\right)+3x\left(t\right)=7$ is $f\left(t\right)=\frac{7}{3}+\frac{7}{3}{e}^{-3t}C$