Q1E - 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 $\mathbf{(}\mathbf{1}\mathbf{+}{\mathbf{t}}^{\mathbf{2}}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{tant}$.

Step-by-Step Solution

Answer

The differential equation has a unique solution.

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

The given differential equation is $(1+{\mathrm{t}}^2)\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}-\mathrm{y}=\mathrm{tant}$.

 

It can be written as $\mathrm{y}\text{'}\text{'}+\frac{\mathrm{t}}{(1+{\mathrm{t}}^2)}\mathrm{y}\text{'}-\frac{\mathrm{y}}{(1+{\mathrm{t}}^2)}=\frac{\mathrm{tant}}{(1+{\mathrm{t}}^2)}$.

 

So, $\mathrm{p}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{(1+{\mathrm{t}}^2)},\mathrm{q}\left(\mathrm{t}\right)=-\frac{1}{(1+{\mathrm{t}}^2)},\mathrm{g}\left(\mathrm{t}\right)=\frac{\mathrm{tant}}{(1+{\mathrm{t}}^2)}$

Step 2: Check the result

Here p(t), q(t), and g(t) are continuous functions in the interval $\mathbf{-}\mathbf{\infty }\mathbf{<}\mathbf{t}\mathbf{<}\mathbf{\infty }$ except at $\frac{\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{\pi }}{\mathbf{2}}$ where n is an integer and here ${\mathbf{t}}_{\mathbf{0}}\mathbf{=}\mathbf{1}$ in the continuity interval.

 

Therefore, the differential equation has a unique solution.