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.

6.$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{,}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

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

Step-by-step solution

Simplification for the differential equation

Consider the equation as follows:

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

Now, separate the variables as follows:

$\begin{array}{c}\frac{dx}{dt}=\frac{1}{x}\\ xdx=dt\end{array}$

Integrating on both sides as follows:

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

Substituting the initial condition as follows:

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

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