Q3E - 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.

${\mathbf{t}}^{\mathbf{2}}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{cost}$

Step-by-Step Solution

Answer

The differential equation has a unique solution in $0<\mathrm{t}<\infty$.

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

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

 

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

 

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

Step 2: Check the result

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

 

Therefore, the differential equation has a unique solution is $0<\mathrm{t}<\infty$.