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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.8.$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}{\mathbf{=}}\sqrt{\mathbf{x}}{\mathbf{,}}{\mathbit{x}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}{\mathbf{=}}{\mathbf{4}}$

The solution is $x\left(t\right)=\frac{{t}^{2}}{4}+4+2t$.

See the step by step solution

## Step 1 : Simplification for the differential equation

Consider the equation as follows.

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

Now, separate the variables as follows.

$\begin{array}{c}\frac{dx}{dt}=\sqrt{x}\\ \frac{dx}{\sqrt{x}}\text{\hspace{0.17em}}=dt\\ {x}^{\frac{-1}{2}}dx=dt\end{array}$

Integrating on both sides as follows.

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

Simplify further as follows:

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

Substituting the initial condition as follows.

$\begin{array}{c}2{x}^{\frac{1}{2}}=t+C\\ 2{\left(4\right)}^{\frac{1}{2}}=0+C\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left\{\because x\left(0\right)=4\right\}\\ 2\left(2\right)=C\\ 4=C\end{array}$

## Step 2 : Calculation of the solution

Now, substitute the value $4$ for $c$ in $2{x}^{\frac{1}{2}}=t+C$ as follows:

$\begin{array}{l}2{x}^{\frac{1}{2}}=t+C\\ 2{x}^{\frac{1}{2}}=t+4\\ {x}^{\frac{1}{2}}=\frac{1}{2}\left(t+4\right)\\ {x}^{\frac{1}{2}}=\frac{t}{2}+\frac{4}{2}\end{array}$

Simplify further as follows.

$\begin{array}{l}{x}^{\frac{1}{2}}=\frac{t}{2}+\frac{4}{2}\\ {x}^{\frac{1}{2}}=\frac{t}{2}+2\end{array}$

Now, squaring on both sides as follows:

$\begin{array}{c}{x}^{\frac{1}{2}}=\frac{t}{2}+2\\ {\left({x}^{\frac{1}{2}}\right)}^{2}={\left(\frac{t}{2}+2\right)}^{2}\\ x=\frac{{t}^{2}}{4}+4+2×2×\frac{t}{2}\\ x\left(t\right)=\frac{{t}^{2}}{4}+4+2t\end{array}$

Hence, the solution for the differential equation$\frac{dx}{dt}=\sqrt{x}$ is $x\left(t\right)=\frac{{t}^{2}}{4}+4+2t$ .