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.

9.$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}{\mathbf{\text{x}}}^{\mathbf{\text{k}}}\mathbf{(}\mathbf{w}\mathbf{i}\mathbf{t}\mathbf{h}\mathbf{k}\mathbf{\ne }\mathbf{1}\mathbf{)}\mathbf{,}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$

Short answer

The solution is$x\left(t\right)={((-k+1)t+1)}^{\frac{1}{-k+1}}$.

Step-by-step solution

:Simplification for the differential equation 

Consider the equation as follows:

$\frac{dx}{dt}=x^k$with$k\ne 0$

Now, separate the variables as follows.

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

Integrating on both sides as follows.

$\begin{array}{c}{x}^{-k}dx=dt\\ \int x^{-k}dx=\int dt\\ \frac{x^{-k+1}}{-k+1}=t+C_1\\ x^{-k+1}=(-k+1)(t+C_1)\end{array}$

Simplify further as follows.

$\begin{array}{l}{x}^{-k+1}=(-k+1)(t+C_1)\\ x^{-k+1}=(-k+1)t+C\{\therefore C=(-k+1)C_1\}\\ x\left(t\right)={((-k+1)t+C)}^{\frac{1}{-k+1}}\end{array}$

Substituting the initial condition as follows:

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

Simplify further as follows.

$\begin{array}{c}{1}^{(-k+1)}=C\\ C=1\end{array}$

Calculation of the solution 

Now, substitute the value $1$ for C in $x\left(t\right)={((-k+1)t+C)}^{\frac{1}{-k+1}}$ as follows.

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

Hence, the answer for the differential equation $\frac{dx}{dt}=x^k$is $x\left(t\right)={((-k+1)t+1)}^{\frac{1}{-k+1}}$