Q24E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 21–26, solve the initial value problem $\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{dt}\mathbf{+}\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{dx}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{x}\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{1}$

Step-by-Step Solution

Answer

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

Step 1: Evaluate the equation is exact

Here $\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{dt}\mathbf{+}\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{dx}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{x}\mathbf{(}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{1}$ 

 

The condition for exact is $\frac{\partial \mathbf{M}}{\partial \mathbf{x}}=\frac{\partial \mathbf{N}}{\partial \mathbf{t}}$  .

 

 $$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{N}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)} \end{gathered}$$

 $\frac{\partial \mathbf{M}}{\partial \mathbf{x}}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\mathbf{=}\frac{\partial \mathbf{N}}{\partial \mathbf{t}}$

 

 

This equation is exact.

Step 2: Find the value of F(x,t)

Here

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{t}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)} \\ \mathbf{F}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\int \mathbf{M}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{dt}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \mathbf{=}\int \mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{dt}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)} \\ \mathbf{=}\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{x}\mathbf{+}\mathbf{t}\mathbf{)}\mathbf{+}\mathbf{g}\mathbf{(}\mathbf{x}\mathbf{)} \end{gathered}$$

Step 3: Determine the value of g(y)

$$\begin{gathered} \frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{N}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)} \\ {\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\mathbf{(}x\mathbf{)}\mathbf{=}\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{1}\mathbf{)} \\ \mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1} \\ \mathbf{g}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{x}\mathbf{+}{\mathbf{C}}_{\mathbf{1}} \end{gathered}$$

Now  $\mathbf{F}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\mathbf{x}\mathbf{+}\mathbf{t}\mathbf{-}\mathbf{x}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}$

 

Therefore, the solution of the differential equation is 

 $$\begin{gathered} {\mathbf{e}}^{\mathbf{t}}\mathbf{x}\mathbf{+}\mathbf{t}\mathbf{-}\mathbf{x}\mathbf{=}\mathbf{C} \\ \mathbf{x}\mathbf{=}\frac{\mathbf{c}\mathbf{-}\mathbf{t}}{{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{1}} \end{gathered}$$

 

Apply the initial conditions$\mathbf{x}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{x}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}$  .

 $$\begin{gathered} \mathbf{1}\mathbf{=}\frac{\mathbf{c}\mathbf{-}\mathbf{1}}{{\mathbf{e}}^{\mathbf{1}}\mathbf{-}\mathbf{1}} \\ \mathbf{C}\mathbf{=}\mathbf{e} \end{gathered}$$

The solutions is  $\mathbf{x}\mathbf{=}\frac{\mathbf{e}\mathbf{-}\mathbf{t}}{{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{1}}$.

 

Hence the solution is  $\mathbf{x}\mathbf{=}\frac{\mathbf{e}\mathbf{-}\mathbf{t}}{{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\mathbf{1}}$