Q4E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1 through 4, use Theorem 5 to discuss the existence and uniqueness of a solution to the differential equation that satisfies the initial conditions $\mathbf{y}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}{\mathbf{Y}}_{\mathbf{o}}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}{\mathbf{Y}}_{\mathbf{1}}$, where ${\mathbf{Y}}_{\mathbf{o}}$ and ${\mathbf{Y}}_{\mathbf{1}}$ are real constants.

${\mathbf{e}}^{\mathbf{t}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{t}\mathbf{-}\mathbf{3}}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{lnt}$

Step-by-Step Solution

Answer

The differential equation has a unique solution in $0<\mathrm{t}<3$.

Step 1: Find the value of p(t),q(t),g(t)

The given differential equation is ${\mathrm{e}}^{\mathrm{t}}\mathrm{y}\text{'}\text{'}-\frac{1}{\mathrm{t}-3}\mathrm{y}\text{'}+\mathrm{y}=\mathrm{lnt}$.

 

It can be written as $\mathrm{y}\text{'}\text{'}-\frac{\mathrm{y}\text{'}}{{\mathrm{e}}^{\mathrm{t}}(\mathrm{t}-3)}+\frac{\mathrm{y}}{{\mathrm{e}}^{\mathrm{t}}(\mathrm{t}-3)}=\frac{\mathrm{lnt}}{{\mathrm{e}}^{\mathrm{t}}(\mathrm{t}-3)}$.

 

So, $\mathrm{p}\left(\mathrm{t}\right)=-\frac{1}{{\mathrm{e}}^{\mathrm{t}}(\mathrm{t}-3)},\mathrm{q}\left(\mathrm{t}\right)=\frac{1}{{\mathrm{e}}^{\mathrm{t}}(\mathrm{t}-3)},\mathrm{g}\left(\mathrm{t}\right)=\frac{\mathrm{lnt}}{{\mathrm{e}}^{\mathrm{t}}(\mathrm{t}-3)}$

Step 2: Check the result

Here p(t),q(t),g(t) are continuous functions in the interval $\mathbf{0}\mathbf{<}\mathbf{t}\mathbf{<}\mathbf{3}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{3}\mathbf{<}\mathbf{t}\mathbf{<}\mathbf{\infty }$ and the point ${\mathbf{t}}_{\mathbf{0}}\mathbf{=}\mathbf{1}$ in the continuity interval $\mathbf{0}\mathbf{<}\mathbf{t}\mathbf{<}\mathbf{3}$

 

Therefore, the differential equation has a unique solution in $0<\mathrm{t}<3$. 

 

This is the required result.