Q7E - 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.

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

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 $\mathrm{y}\text{'}\text{'}+\mathrm{ty}\text{'}-{\mathrm{t}}^2\mathrm{y}=0$.

 

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

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),g(t) is a continuous function in the interval $-\infty <\mathrm{t}<\infty$,but initial conditions are $\mathrm{y}\left(0\right)=1,\mathrm{y}\left(1\right)=0$.So it is not an initial value problem.

 

Therefore, the differential equation has no unique solution.

 

This is the required result.