Question

In Problems 9–20, determine whether the equation is exact.

If it is, then solve it.

${\mathbf{\text{e}}}^{\mathbf{\text{t}}}\left(\text{y - t}\right)\mathbf{\text{dt +}}\left({\text{1 +e}}^{\text{t}}\right)\mathbf{\text{dy = 0}}$

Short answer

The solution is$\mathbf{y}\mathbf{=}\frac{\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{C}}{\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{t}}}$.

Step-by-step solution

Evaluate whether the equation is exact

Here${\text{e}}^{\text{t}}\left(\text{y - t}\right)\text{dt +}\left({\text{1 +e}}^{\text{t}}\right)\text{dy = 0}$

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

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

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

This equation is exact.

Find the value of F (x, t)

Here

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

Determine the value of g(y)

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

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

The general solution of the differential equation is$\mathbf{y}\mathbf{=}\frac{\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{C}}{\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{t}}}$

Hence the solution isrole="math" localid="1664172391787" $\mathbf{y}\mathbf{=}\frac{\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{1}\mathbf{)}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\mathbf{C}}{\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{t}}}$