Q5E - 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.

${\mathbf{t}}^{\mathbf{2}}\mathbf{z}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{tz}\mathbf{\text{'}}\mathbf{+}\mathbf{z}\mathbf{=}\mathbf{cost}\mathbf{;}\mathbf{z}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{z}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

The differential equation has no unique solution in $-\infty <\mathrm{t}<\infty$.

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

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

 

It can be written as $\mathrm{z}\text{'}\text{'}+\frac{1}{\mathrm{t}}\mathrm{z}\text{'}+\frac{1}{{\mathrm{t}}^2}\mathrm{z}=\frac{\mathrm{cost}}{{\mathrm{t}}^2}$.

 

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

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),q(t), and g(t) are continuous functions in the interval $0<\mathrm{t}<\infty$, and the point ${\mathrm{t}}_0=0$ isn't in the interval $-\infty <\mathrm{t}<\infty$

 

Therefore, the differential equation has no unique solution in $-\infty <\mathrm{t}<\infty$. 

 

This is the required result.