Question

Solve the nonlinear differential equations in Exercises 6through 11 using the method of separation of variables:Write the differential equation $\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{f}\mathit{x}$as $\frac{\mathbf{d}\mathbf{x}}{\mathbf{f}\mathbf{x}}\mathbf{=}\mathit{d}\mathit{t}$and integrate both sides.

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

Short answer

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

Step-by-step solution

Simplification for the differential equation

Consider the equation as follows.

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

Now, separate the variables as follows.

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

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:

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

Substituting the initial condition as follows.

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

Calculation of the solution 

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

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

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:

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

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