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.

10.$\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\mathbf{,}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

The solution is$x\left(t\right)=si{n}^{-1}t$.

Step-by-step solution

Explanation for the differential equation. 

Consider the equation as follows:

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

Now, separate the variables as follows:

$\begin{array}{c}\frac{dx}{dt}=\frac{1}{cosx}\\ cosxdx=dt\end{array}$

Integrating on both sides as follows:

$\begin{array}{r}cosxdx=dt\\ \int cosxdx=\int dt\\ sinx=t+C\end{array}$

Substituting the initial condition as follows.

$\begin{array}{c}sinx=t+C\\ sinx\left(0\right)=0+C\\ 0=0+C\\ C=0\end{array}$

Calculation of the solution 

Now, substitute the value$0$ for C in $sinx=t+C$ as follows.

$\begin{array}{l}sinx=t+C\\ sinx=t+0\\ sinx=t\\ x\left(t\right)=si{n}^{-1}t\end{array}$

Hence, the solution for the differential equation $\frac{dx}{dt}=\frac{1}{cosx}$is $x\left(t\right)=si{n}^{-1}t$.