Q14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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

If it is, then solve it.

$\left(\frac{t}{y}\right)\mathbf{dy}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{ln}\mathbf{ }\mathbf{y}\mathbf{)}\mathbf{dt}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The solution is $\mathbf{t} \mathbf{ln} \mathbf{y}\mathbf{+}\mathbf{t}\mathbf{=}\mathbf{C}$.

Step 1: Evaluate whether the equation is exact

Here $\left(\frac{\mathbf{t}}{\mathbf{y}}\right)\mathbf{dy}\mathbf{+}\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{ln} \mathbf{y}\mathbf{)}\mathbf{dt}\mathbf{=}\mathbf{0}$

 

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

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{t}}{\mathbf{y}} \\ \mathbf{N}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{ln} \mathbf{y} \end{gathered}$$

$\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{y}}\mathbf{=}\frac{\partial \mathbf{N}}{\partial t}$ 

This equation is exact.

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

Here

 

$$\begin{gathered} \mathbf{M}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\frac{\mathbf{t}}{\mathbf{y}} \\ \mathbf{F}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{=}\int \mathbf{M}\mathbf{(}\mathbf{y}\mathbf{,}\mathbf{t}\mathbf{)}\mathbf{dy}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{=}\int \frac{\mathbf{t}}{\mathbf{y}}\mathbf{dy}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \\ \mathbf{=}\mathbf{t} \mathbf{ln} \mathbf{y}\mathbf{+}\mathbf{g}\mathbf{\left(}\mathbf{t}\mathbf{\right)} \end{gathered}$$

Step 3: determine the value of g (t)

Now  $\text{F}\left(\text{y,t}\right)\text{ = t} \text{ln} {\text{y + t + C}}_{\text{1}}$

 

The general solution of the differential equation is $\mathbf{t} \mathbf{ln} \mathbf{y}\mathbf{+}\mathbf{t}\mathbf{=}\mathbf{C}$.

 

Hence the solution is $\mathbf{t} \mathbf{ln} \mathbf{y}\mathbf{+}\mathbf{t}\mathbf{=}\mathbf{C}$