Q2E - 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.

$\mathbf{t}\mathbf{(}\mathbf{t}\mathbf{-}\mathbf{3}\mathbf{)}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{ty}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}$

Step-by-Step Solution

Answer

The differential equation has a unique solution in 0<t<3.

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

The given differential equation is $\mathrm{t}(\mathrm{t}-3)\mathrm{y}\text{'}\text{'}+2\mathrm{ty}\text{'}-\mathrm{y}={\mathrm{t}}^2$

 

It can be written as $\mathrm{y}\text{'}\text{'}+\frac{2\mathrm{t}}{\mathrm{t}(\mathrm{t}-3)}\mathrm{y}\text{'}-\frac{\mathrm{y}}{\mathrm{t}(\mathrm{t}-3)}=\frac{{\mathrm{t}}^2}{\mathrm{t}(\mathrm{t}-3)}$.

 

So, $\mathrm{p}\left(\mathrm{t}\right)=\frac{2\mathrm{t}}{\mathrm{t}(\mathrm{t}-3)},\mathrm{q}\left(\mathrm{t}\right)=-\frac{1}{\mathrm{t}(\mathrm{t}-3)},\mathrm{g}\left(\mathrm{t}\right)=\frac{{\mathrm{t}}^2}{\mathrm{t}(\mathrm{t}-3)}$

Step 2: Check the result

Here p(t), q(t),g(t) are continuous functions in the interval 0<t<3 and $\mathbf{2}\mathbf{<}\mathbf{t}\mathbf{<}\mathbf{\infty }$ and the point ${\mathbf{t}}_{\mathbf{0}}\mathbf{=}\mathbf{1}$ in the continuity interval 0<t<3.

 

Therefore, the differential equation has a unique solution in 0<t<3.