Q23E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 21–26, solve the initial value problem.

$\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{y}\mathbf{+}{\mathbf{te}}^{\mathbf{t}}\mathbf{y}\mathbf{)}\mathbf{dt}\mathbf{+}\mathbf{(}\mathbf{t}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

The solution is .$\mathbf{y}\mathbf{=}\frac{\mathbf{-}\mathbf{2}}{{\mathbf{te}}^{\mathbf{t}}\mathbf{+}\mathbf{2}}$

Step 1: Evaluate the equation is exact

Here $\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{y}\mathbf{+}{\mathbf{te}}^{\mathbf{t}}\mathbf{y}\mathbf{)}\mathbf{dt}\mathbf{+}\mathbf{(}\mathbf{t}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{2}\mathbf{)}\mathbf{dy}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{(}\mathbf{0}\mathbf{)}\mathbf{=}\mathbf{-}\mathbf{1}$

 

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

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{t}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{(}{\mathbf{e}}^{\mathbf{t}}\mathbf{y}\mathbf{+}{\mathbf{te}}^{\mathbf{t}}\mathbf{y}\mathbf{)} \\ \mathbf{N}\mathbf{(}\mathbf{t}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{(}\mathbf{t}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{2}\mathbf{)} \\ \frac{\mathbf{\partial }\mathbf{M}}{\mathbf{\partial }\mathbf{y}}\mathbf{=}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}{\mathbf{te}}^{\mathbf{t}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{N}}{\mathbf{\partial }\mathbf{t}} \end{gathered}$$

This equation is exact.

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

Here

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

Step 3: Determine the value of g(y)

$$\begin{gathered} \frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{(}t\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{N}\mathbf{(}t\mathbf{,}\mathbf{y}\mathbf{)} \\ {\mathbf{te}}^{\mathbf{t}}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{(}\mathbf{t} {\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{2}\mathbf{)} \\ \mathbf{g}\mathbf{\text{'}}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}2 \\ \mathbf{g}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{y}\mathbf{+}{\mathbf{C}}_{\mathbf{1}} \end{gathered}$$

Now$\mathbf{F}\mathbf{(}\mathbf{t}\mathbf{,}\mathbf{y}\mathbf{)}\mathbf{=}\mathbf{yt}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{+}{\mathbf{C}}_{\mathbf{1}}$

 

Therefore the solution of the differential equation is 

 $$\begin{gathered} {\mathbf{yte}}^{\mathbf{t}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{C} \\ \mathbf{y}\mathbf{=}\frac{\mathbf{c}}{{\mathbf{te}}^{\mathbf{t}}\mathbf{+}\mathbf{2}} \end{gathered}$$

 Apply the initial conditions.

 $\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}$

$$\begin{gathered} \mathbf{-}\mathbf{1}\mathbf{=}\frac{\mathbf{c}}{\mathbf{0}\mathbf{+}\mathbf{2}} \\ \mathbf{C}\mathbf{=}\mathbf{-}\mathbf{2} \end{gathered}$$

 The solutions is$\mathbf{y}\mathbf{=}\frac{\mathbf{-}\mathbf{2}}{{\mathbf{te}}^{\mathbf{t}}\mathbf{+}\mathbf{2}}$ .

 

Hence the solution is  $\mathbf{y}\mathbf{=}\frac{\mathbf{-}\mathbf{2}}{{\mathbf{te}}^{\mathbf{t}}\mathbf{+}\mathbf{2}}$