Q39E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 38 and 39, use the elimination method of Section to find a general solution to the given system.

$$\begin{gathered} {\mathit{d}}^{\mathbf{2}}\mathit{x}\mathbf{/}\mathit{d}{\mathit{t}}^{\mathbf{2}}\mathbf{-}\mathit{x}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathit{x}\mathbf{+}{\mathit{d}}^{\mathbf{2}}\mathit{y}\mathbf{/}\mathit{d}{\mathit{t}}^{\mathbf{2}}\mathbf{-}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{3}\mathbf{t}} \end{gathered}$$

Step-by-Step Solution

Answer

The general solution is $$\begin{gathered} x\left(t\right)=c_1+c_{2}t+c_3{e}^{\sqrt{2}t}+c_4{e}^{-\sqrt{2}t}-\frac{1}{63}{e}^{-3t} \\ y\left(t\right)=c_1+c_{2}t-c_3{e}^{\sqrt{2}t}-c_4{e}^{-\sqrt{2}t}+\frac{8}{63}{e}^{-3t} \end{gathered}$$

Step 1: Definition

A differential equation is an equation that contains one or more functions with its derivatives.

Step 2: Simplify equation

It is given that:

$$\begin{gathered} x^{\text{'}\text{'}}-x+y=0 \\ x+y^{\text{'}\text{'}}-y=e^{3t} \end{gathered}$$

Writing in operator form we have:

$$\begin{gathered} \left(D^2-1\right)\left[x\right]+y=0 \\ x+\left(D^2-1\right)\left[y\right]=e^{3t} \end{gathered}$$

Solve for $y: equation \left(1\right)-\left(D^2-1\right)\times equation \left(2\right)$

This gives:

$$\begin{gathered} \left[{\left(D^2-1\right)}^2-1\right]\left[x\right]=0-e^{3t} \\ \left(D^4-2D^2+1-1\right)\left[x\right]=-e^{3t} \\ \left(D^4-2D^2\right)\left[x\right]=-e^{3t} \\ {D}^2\left(D^2-2\right)\left[x\right]=-e^{3t} \end{gathered}$$

So, the general equation is: 

$$\begin{gathered} D^2-1=1,-1 \\ {D}^2=2,0 \\ {y}_g=C_1+C_{2}t+C_3{e}^{\sqrt{2}t}+C_4{e}^{-\sqrt{2}t} \end{gathered}$$

Step 3: For particular solution

Let particular solution be $y_p=k{e}^{3t}$

$$\begin{gathered} y_p^{\text{'}}=3k{e}^{3t} \\ {y}_p^{\text{'}\text{'}}=9k{e}^{3t} \\ {y}_p^{\text{'}\text{'}\text{'}}=27k{e}^{3t} \\ {y}_p^{\left(4\right)}=81k{e}^{3t} \end{gathered}$$

$$\begin{gathered} \left(2D^2-D^4\right)y=-8e^{3t} \\ 2y^{\text{'}\text{'}}-y^{\left(4\right)}=-8e^{3t} \\ 18k{e}^{3t}-81k{e}^{3t}=-8e^{3t} \\ \Rightarrow k=\frac{8}{63} \\ \Rightarrow y_p=\frac{8}{63}{e}^{3t} \end{gathered}$$

So  $y=C_1+C_{2}t+C_3{e}^{\sqrt{2}t}+C_4{e}^{-\sqrt{2}t}+\frac{8e^{3t}}{63}$

Substituting values in $x=y-y\text{'}\text{'}+e^{3t}$we get:

$$\begin{gathered} x=y-y^{\text{'}\text{'}}+e^{3t} \\ =C_1+C_{2}t+C_3{e}^{\sqrt{2}t}+C_4{e}^{-\sqrt{2}t}+\frac{8}{63}{e}^{3t}+e^{3t}-2C_3{e}^{\sqrt{2}t}-2C_4{e}^{-\sqrt{2}t}-\frac{8}{7}{e}^{3t}x \\ =C_1+C_{2}t-C_3{e}^{\sqrt{2}t}-C_4{e}^{-\sqrt{2}t}-\frac{e^{3t}}{63} \end{gathered}$$

Therefore the solution of the system is:$$\begin{gathered} x\left(t\right)=c_1+c_{2}t+c_3{e}^{\sqrt{2}t}+c_4{e}^{-\sqrt{2}t}-\frac{1}{63}{e}^{-3t} \\ y\left(t\right)=c_1+c_{2}t-c_3{e}^{\sqrt{2}t}-c_4{e}^{-\sqrt{2}t}+\frac{8}{63}{e}^{-3t} \end{gathered}$$