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.

7.$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathbf{\text{x}}}^{\mathbf{2}}\mathbf{,}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Describe the behavior of your solution as t increases.

Short answer

The solution is$x=\frac{1}{1-t}$

Step-by-step solution

Simplification for the differential equation

Consider the equation as follows.

$\frac{dx}{dt}=x^2$

Now, separate the variables as follows.

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

Integrating on both sides as follows.

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

Substituting the initial condition as follows.

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

Calculation of the solution

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

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

As the relation between $t$ and $x$ is inverse and then$x$ will decreases when $t$ increases.

The solution becomes as follows:

$x=\frac{1}{1-t}$

Hence, the solution for the differential equation$\frac{dx}{dt}=x^2$ is $x=\frac{1}{1-t}$.