Q16E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 3 – 18, use the elimination method to find a general solution for the given linear system, where differentiation is with respect to t.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{x}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{,} \\ \mathbf{2}\mathbf{x}\mathbf{+}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{0} \end{gathered}$$

Step-by-Step Solution

Answer

The solutions for the given linear system are $\mathrm{x}\left(\mathrm{t}\right)={\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{t}}+\frac{2}{9}{\mathrm{e}}^{4\mathrm{t}}$ and $\mathrm{y}\left(\mathrm{t}\right)=-\frac{1}{2}{\mathrm{c}}_1{\mathrm{e}}^{-2\mathrm{t}}-2{\mathrm{c}}_2{\mathrm{e}}^{\mathrm{t}}-\frac{1}{36}{\mathrm{e}}^{4\mathrm{t}}+{\mathrm{c}}_3$.

Step 1: General form

Elimination Procedure for 2 × 2 Systems

 

To find a general solution for the system

$$\begin{gathered} {\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,} \\ {\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,} \end{gathered}$$ 

 Where ${\mathbf{L}}_{\mathbf{1}}\mathbf{,}{\mathbf{L}}_{\mathbf{2}}\mathbf{,}{\mathbf{L}}_{\mathbf{3}}\mathbf{,}$ and ${\mathbf{L}}_{\mathbf{4}}$ are polynomials in $\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$

 

1. Make sure that the system is written in operator form.
    
    
2. Eliminate one of the variables, say, y, and solve the resulting equation for x(t). If the system is degenerating, stop! A separate analysis is required to determine whether or not there are solutions.

 

1. (Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), and y(t) give the desired general solution.

 

1. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]

 

1. Remove the extra constants by substituting the expressions for x(t) and y(t) into one or both of the equations in the system. Write the expressions for x(t) and y(t) in terms of the remaining constants.

Step 2: Evaluate the given equation

Given that,

 

 $\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{+}\mathbf{x}\mathbf{+}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}$     … (1)

 

   $\mathbf{2}\mathbf{x}\mathbf{+}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{{\mathbf{dt}}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$         … (2)

 

Let us rewrite the system in operator form,

 

 $\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}\mathbf{D}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}$        … (3)

 $2\mathbf{x}\mathbf{+}{\mathbf{D}}^2\left[\mathbf{y}\right]\mathbf{=}0$                  … (4)

 

Multiply D on both sides of equation (3) then subtract equation (3) and (4) together one gets,

$$\begin{gathered} \mathbf{D}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}{\mathbf{D}}^{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{-}\left(\mathbf{2}\mathbf{x}\mathbf{+}{\mathbf{D}}^{\mathbf{2}}\left[\mathbf{y}\right]\right)\mathbf{=}\mathbf{D}\left[{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\right] \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[\mathbf{x}\right]\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \dots \left(5\right) \end{gathered}$$ 

 Since the auxiliary equation to the corresponding homogeneous equation is ${\mathbf{r}}^{\mathbf{2}}\mathbf{+}\mathbf{r}\mathbf{-}\mathbf{2}\mathbf{=}\mathbf{0}$. The roots are r=1 and r=-2 .

 

Then, the homogeneous solution of u is;

 

  ${\mathbf{x}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}$      … (6)

 

Let us take the undetermined coefficients and assume that,

 

${\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}$                … (7)

 

Now find the derivate the equation (7).

$$\begin{gathered} \mathbf{D}\left[{\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right)\right]\mathbf{=}\mathbf{4}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}} \\ {\mathbf{D}}^{\mathbf{2}}\left[{\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right)\right]\mathbf{=}\mathbf{16}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}} \end{gathered}$$ 

Step 3: Substitution method

Substitute the derivation in equation (5).

 $$\begin{gathered} \left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}\right]\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \mathbf{16}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}\mathbf{+}\mathbf{4}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \mathbf{18}{\mathbf{Ae}}^{\mathbf{4}\mathbf{t}}\mathbf{=}\mathbf{4}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \end{gathered}$$

 Now, equalize the like terms.

$$\begin{gathered} \mathbf{18}\mathbf{A}\mathbf{=}\mathbf{4} \\ \mathbf{A}\mathbf{=}\frac{\mathbf{4}}{\mathbf{18}} \\ \mathbf{A}\mathbf{=}\frac{\mathbf{2}}{\mathbf{9}} \end{gathered}$$ 

So, ${\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right)\mathbf{=}\frac{2}{9}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}$      … (8)

 

Use equations (6) and (8) to get,

 $$\begin{gathered} \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{x}}_{\mathbf{h}}\left(\mathbf{t}\right)\mathbf{+}{\mathbf{x}}_{\mathbf{p}}\left(\mathbf{t}\right) \\ \mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\frac{2}{9}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \dots \left(9\right) \end{gathered}$$

 Now, take equation (3).

$$\begin{gathered} \left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right]\mathbf{+}\mathbf{D}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \mathbf{D}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[\mathbf{x}\right] \\ \mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{-}\left(\mathbf{D}\mathbf{+}\mathbf{1}\right)\left[{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{9}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\right] \\ \mathbf{=}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{+}\mathbf{2}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{8}}{\mathbf{9}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{2}}{\mathbf{9}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{9}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}} \\ \mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{36}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}} \end{gathered}$$ 

 Thus, the solutions for the given linear system are $\mathbf{x}\left(\mathbf{t}\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{+}\frac{2}{9}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}$ and $\mathbf{y}\left(\mathbf{t}\right)\mathbf{=}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{c}}_{\mathbf{1}}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{t}}\mathbf{-}\mathbf{2}{\mathbf{c}}_{\mathbf{2}}{\mathbf{e}}^{\mathbf{t}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{36}}{\mathbf{e}}^{\mathbf{4}\mathbf{t}}\mathbf{+}{\mathbf{c}}_{\mathbf{3}}$.