Q48E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

All that is known concerning a mysterious second-order constant-coefficient differential equation $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{py}\mathbf{\text{'}}\mathbf{+}\mathbf{qy}\mathbf{=}\mathbf{g}\left(\mathbf{t}\right)$ is that  ${\mathbf{t}}^2\mathbf{+}\mathbf{1}\mathbf{+}{\mathbf{e}}^t\mathbf{cost}\mathbf{,}{\mathbf{t}}^2\mathbf{+}\mathbf{1}\mathbf{+}{\mathbf{e}}^t\mathbf{sint}$ and ${\mathbf{t}}^2\mathbf{+}\mathbf{1}\mathbf{+}{\mathbf{e}}^t\mathbf{cost}\mathbf{+}{\mathbf{e}}^t\mathbf{sint}$  are solutions. 

 

(a)Determine two linearly independent solutions to the corresponding homogeneous equation. 

 

(b) Find a suitable choice of p, q, and g(t) that enables these solutions.

Step-by-Step Solution

Answer

a.$\mathbf{y}\mathbf{=}{\mathbf{e}}^t\mathbf{cost}$ and $\mathbf{y}\mathbf{=}{\mathbf{e}}^t\mathbf{sint}$

b.$\mathbf{p}\mathbf{=}\mathbf{-}\mathbf{2}$,$\mathbf{q}\mathbf{=}\mathbf{2}$ and  $\mathbf{g}\left(\mathbf{t}\right)\mathbf{=}\mathbf{2}{\mathbf{t}}^2\mathbf{-}\mathbf{4}\mathbf{t}\mathbf{+}\mathbf{4}$

Step 1: Determine first linearly independent solutions to the corresponding homogeneous equation.

The differential equation is,

 

$\mathrm{y}\text{'}\text{'}+\mathrm{py}\text{'}+\mathrm{qy}=\mathrm{g}\left(\mathrm{t}\right)......\left(1\right)$ 

 

Write the homogeneous differential equation of the equation (1),

 

 $\mathrm{y}\text{'}\text{'}+\mathrm{py}\text{'}+\mathrm{qy}=0......\left(2\right)$

 

Let the solution for the above equation,

 

$\mathrm{y}={\mathrm{e}}^{\mathrm{t}}\mathrm{cost}$

Now find the first and second derivatives of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\mathrm{cost}-{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\mathrm{cost}-{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}-({\mathrm{e}}^{\mathrm{t}}\mathrm{sint}+{\mathrm{e}}^{\mathrm{t}}\mathrm{cost})\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=-2{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}\end{array}$

Substitute the value of ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$   and  ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$  in the equation (2),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+\mathrm{py}\text{'}+\mathrm{qy}=0\\ -2{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}+\mathrm{p}[{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}-{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}]+\mathrm{q}\left[{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}\right]=0\\ -[2+\mathrm{p}]{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}+[\mathrm{p}+\mathrm{q}]{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}=0\end{array}$

Comparing all coefficients of the above equation,

 $\begin{array}{c}-[2+\mathrm{p}]=0\Rightarrow \mathrm{p}=-2\\ \mathrm{p}+\mathrm{q}=0 ......\left(3\right)\end{array}$

Substitute the value of p in the equation (3),

 $\begin{array}{c}\mathrm{p}+\mathrm{q}=0\\ -2+\mathrm{q}=0\\ \mathrm{q}=2\end{array}$

Substitute the value of p and q in the equation (2),

 $\begin{array}{c}\mathrm{y}\text{'}\text{'}+\mathrm{py}\text{'}+\mathrm{qy}=0\\ \mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+2\mathrm{y}=0\end{array}$

Therefore, the particular solution of equation (1),

$\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+2\mathrm{y}=0$

Step 2: Determine second linearly independent solutions to the corresponding homogeneous equation.

The differential equation is,

 

$\mathrm{y}\text{'}\text{'}+\mathrm{py}\text{'}+\mathrm{qy}=\mathrm{g}\left(\mathrm{t}\right)......\left(1\right)$ 

 

Write the homogeneous differential equation of the equation (1),

 

 $\mathrm{y}\text{'}\text{'}+\mathrm{py}\text{'}+\mathrm{qy}=0 ......\left(2\right)$

 

Let the solution for the above equation,

$\mathrm{y}={\mathrm{e}}^{\mathrm{t}}\mathrm{sint}$

Now find the first and second derivatives of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\mathrm{sint}+{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{t}}\mathrm{sint}+{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}+({\mathrm{e}}^{\mathrm{t}}\mathrm{cost}-{\mathrm{e}}^{\mathrm{t}}\mathrm{sint})\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=2{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}\end{array}$

Substitute the value of  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$  and  ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$  in the equation (2),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+\mathrm{py}\text{'}+\mathrm{qy}=0\\ 2{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}+\mathrm{p}[{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}+{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}]+\mathrm{q}\left[{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}\right]=0\\ [2+\mathrm{p}]{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}+[\mathrm{p}+\mathrm{q}]{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}=0\end{array}$

Comparing all coefficients of the above equation,

$\begin{array}{c}2+\mathrm{p}=0\Rightarrow \mathrm{p}=-2\\ \mathrm{p}+\mathrm{q}=0 ......\left(3\right)\end{array}$

Substitute the value of p in the equation (3),

 $\begin{array}{c}\mathrm{p}+\mathrm{q}=0\\ -2+\mathrm{q}=0\\ \mathrm{q}=2\end{array}$

 

 

Substitute the value of p and q in the equation (2),

 $\begin{array}{c}\mathrm{y}\text{'}\text{'}+\mathrm{py}\text{'}+\mathrm{qy}=0\\ \mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+2\mathrm{y}=0\end{array}$

 

 

Hence, the particular solution of equation (1),

 

 $\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+2\mathrm{y}=0$

 

The two linearly independent solutions to the corresponding homogeneous equation are,

 $\mathrm{y}={\mathrm{e}}^{\mathrm{t}}\mathrm{cost}$  and $\mathrm{y}={\mathrm{e}}^{\mathrm{t}}\mathrm{sint}$

Step 3: Find a value of p, q, and g(t).

From the step 1,

 

$\mathrm{p}=-2$  and  $\mathrm{q}=2$

Substitute the value of p and q in the equation (1),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}+\mathrm{py}\text{'}+\mathrm{qy}=\mathrm{g}\left(\mathrm{t}\right)\\ \mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+2\mathrm{y}=\mathrm{g}\left(\mathrm{t}\right)......\left(4\right)\end{array}$

Let a particular solution,

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^2+1$

Now find the first and second derivatives of the above equation,

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=2\mathrm{t}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=2\end{array}$

Substitute the value of  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$   in the equation (4),

$\begin{array}{c}\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+2\mathrm{y}=\mathrm{g}\left(\mathrm{t}\right)\\ 2-2\left(2\mathrm{t}\right)+2({\mathrm{t}}^2+1)=\mathrm{g}\left(\mathrm{t}\right)\\ 2{\mathrm{t}}^2-4\mathrm{t}+4=\mathrm{g}\left(\mathrm{t}\right)\\ \mathrm{g}\left(\mathrm{t}\right)=2{\mathrm{t}}^2-4\mathrm{t}+4\end{array}$