Q6E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 5 through 8, determine whether Theorem 5 applies. If it does, then discuss what conclusions can be drawn. If it does not, explain why.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{yy}\mathbf{\text{'}}\mathbf{=}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{1}\mathbf{;}\mathbf{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

The differential equation has no unique solution.

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

The given differential equation is ${\mathrm{t}}^2\mathrm{z}\text{'}\text{'}+\mathrm{tz}\text{'}+\mathrm{z}=\mathrm{cost}$.

So, $\mathrm{p}\left(\mathrm{t}\right)=\mathrm{y},\mathrm{q}\left(\mathrm{t}\right)=0,\mathrm{g}\left(\mathrm{t}\right)={\mathrm{t}}^2-1$

Step 2: Check the result.

From theorem (5) If p(t), q(t), and g(t) are continuous on an interval (a, b) that contains the point t, then for any choice of the initial values ${\mathbf{Y}}_{\mathbf{o}} \mathbf{and} {\mathbf{Y}}_{\mathbf{1}}$, there exists a unique solution y(1) on the same interval (a, b) to the initial value problems.

 

Here p(t) is a function of y, so one can not apply theorem (5).

 

Therefore, the differential equation has no unique solution.

 

This is the required result.