Question

Question: In Problems 1 - 30, solve the equation.

$\frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{cos}}^2\left(t-x\right)$

Short answer

The solution of the given equation is $\mathbf{tan}\left(t-x\right)+\mathbf{t}=\mathbf{C}$.

Step-by-step solution

Given information and simplification

Given that, $\frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{cos}}^2\left(t-x\right)\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Evaluate the equation (1).

Let us take$\mathbf{u}\mathbf{=}\mathbf{t}\mathbf{-}\mathbf{x}$. Then, $\frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\mathbf{1}\mathbf{-}\frac{\mathrm{du}}{\mathrm{dt}}$.

$$\begin{gathered} \begin{array}{c}\frac{\mathrm{dx}}{\mathrm{dt}}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{cos}}^2\left(t-x\right)\\ \mathbf{1}\mathbf{-}\frac{\mathrm{du}}{\mathrm{dt}}\mathbf{=}\mathbf{1}\mathbf{+}{\mathbf{cos}}^2\mathbf{u}\\ \mathbf{-}\frac{\mathrm{du}}{\mathrm{dt}}\mathbf{=}{\mathbf{cos}}^2\mathbf{u}\\ \mathbf{-}\frac{1}{{\mathbf{cos}}^2\mathbf{u}}\mathbf{du}\mathbf{=}\mathbf{dt}\end{array} \\ \mathbf{-}\frac{1}{{\mathbf{cos}}^2\mathbf{u}}\mathbf{du}\mathbf{=}\mathbf{dt}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right) \end{gathered}$$

Integration                    

Now integrate the equation (2) on both sides.

$\begin{array}{c}\mathbf{-}\int \frac{1}{{\mathbf{cos}}^2\mathbf{u}}\mathbf{du}\mathbf{=}\int \mathbf{dt}\\ \mathbf{-}\int {\mathbf{sec}}^2\mathbf{u}\mathbf{du}\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{C}\\ \mathbf{-}\mathbf{tan}\mathbf{u}\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{C}\end{array}$

Substitute the value of u here.

$\mathbf{tan}\left(t-x\right)+\mathbf{t}=\mathbf{C}$

So, the solution is$\mathbf{tan}\left(t-x\right)+\mathbf{t}=\mathbf{C}$