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.

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

Short answer

The differential equation has a unique solution.

Step-by-step solution

Find the value of p(t),q(t),g(t)

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

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

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

Check the result

From theorem (5) If p(t), q(t), and g(t) are continuous on an interval (a, b) that contains 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 but shows discontinuity at t=1.

So, theorem (5) applies except for point t=1.

Since the initial conditions are not the point (1,0) but are at (0,1).

Therefore the differential equation has a unique solution.s

This is the required result.