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Q6E

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Found in: Page 425

### Linear Algebra With Applications

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

# Solve the nonlinear differential equations in Exercises 6through 11 using the method of separation of variables:Write the differential equation $\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}{\mathbit{f}}{\mathbit{x}}$as$\frac{\mathbf{d}\mathbf{x}}{\mathbf{f}\mathbf{x}}{\mathbf{=}}{\mathbit{d}}{\mathbit{t}}$ and integrate both sides.6.$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{x}}{\mathbf{,}}{\mathbit{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}{\mathbf{1}}$

The solution is $x\left(t\right)=\sqrt{2t+1}$

See the step by step solution

## Step 1: Simplification for the differential equation

Consider the equation as follows:

$\frac{dx}{dt}=\frac{1}{x}$

Now, separate the variables as follows:

$\begin{array}{c}\frac{dx}{dt}=\frac{1}{x}\\ x\text{\hspace{0.17em}}dx=dt\end{array}$

Integrating on both sides as follows:

$\begin{array}{c}x\text{\hspace{0.17em}}dx=dt\\ \int x\text{\hspace{0.17em}}dx\text{\hspace{0.17em}}=\text{\hspace{0.17em}}\int dt\\ \frac{{x}^{2}}{2}=t+C\end{array}$

Substituting the initial condition as follows:

$\begin{array}{c}\frac{{x}^{2}}{2}=t+C\\ \frac{{\left(1\right)}^{2}}{2}=0+C\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left\{Qx\left(0\right)=1\right\}\\ \frac{1}{2}=C\end{array}$

## Step 2 : Calculation of the solution

Now, substitute the value $\frac{1}{2}$ for C in $\frac{{x}^{2}}{2}=t+C$ as follows:

$\begin{array}{c}\frac{{x}^{2}}{2}=t+C\\ \frac{{x}^{2}}{2}=t+\frac{1}{2}\\ \frac{{x}^{2}}{2}=\frac{2t}{2}+\frac{1}{2}\\ \frac{{x}^{2}}{2}=\frac{2t+1}{2}\end{array}$

Simplify further as follows:

$\begin{array}{c}\frac{{x}^{2}}{2}=\frac{2t+1}{2}\\ {x}^{2}=2t+1\\ x=\sqrt{2t+1}\\ x\left(t\right)=\sqrt{2t+1}\end{array}$

Hence, the solution for the differential equation$\frac{dx}{dt}=\frac{1}{x}$ is $x\left(t\right)=\sqrt{2t+1}$