Q38E - 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.

$\mathit{x}\mathbf{-}{\mathit{d}}^{\mathbf{2}}\mathit{y}\mathbf{/}\mathit{d}{\mathit{t}}^{\mathbf{2}}\mathbf{=}\mathit{t}\mathbf{+}\mathbf{1}$

$\mathit{d}\mathit{x}\mathbf{/}\mathit{d}\mathit{t}\mathbf{+}\mathit{d}\mathit{y}\mathbf{/}\mathit{d}\mathit{t}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{=}{\mathit{e}}^{\mathbf{t}}$

Step-by-Step Solution

Answer

The general solution is $$\begin{gathered} x\left(t\right)=\frac{-3c_1-\sqrt{7}{c}_2}{2}{e}^{-\frac{t}{2}}\cos \frac{\sqrt{7}}{2}t+\frac{\sqrt{7}{c}_1-3c_2}{2}{e}^{-\frac{t}{2}}\sin \frac{\sqrt{7}}{2}t+\left(c_3+\frac{1}{2}\right)e^t+\frac{1}{4}t{e}^t+t+1 \\ y\left(t\right)=c_1{e}^{-\frac{t}{2}}\cos \frac{\sqrt{7}}{2}t+c_2{e}^{-\frac{t}{2}}\sin \frac{\sqrt{7}}{2}t+c_3{e}^t+\frac{1}{4}t{e}^t+1 \end{gathered}$$

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 $x-y^{\text{'}\text{'}}=t+1$

$\Rightarrow x=y^{\text{'}\text{'}}+t+1$-------(1)

Differentiating equation (1) we have:

$x^{\text{'}}=y^{\text{'}\text{'}\text{'}}+1$

Substituting value we get:

$x^{\text{'}}+y^{\text{'}}-2y=e^t\Rightarrow y^{\text{'}\text{'}\text{'}}+y^{\text{'}}-2y+1=e^{\text{'}}$

Step 3: For general solution

The auxiliary equation is given by:

$\left(D^3+D-2\right)y=e^t-1$

The homogenous equation is $D^3+D-2=0$:

$D=1,\frac{-1}{2}\pm i\frac{\sqrt{7}}{2}$

So, the general solution is given by $y_g=C_1{e}^t+C_2{e}^{\frac{-t}{2}}\cos \left(\frac{\sqrt{7}}{2}t\right)+C_3{e}^{\frac{-t}{2}}\sin \left(\frac{\sqrt{7}t}{2}\right)$

Step 4: For particular solution

Let $y_p=At{e}^t+B$

Differentiating we have:

$$\begin{gathered} y_p^{\text{'}}=A{e}^t(t+1) \\ {y}_p^{\text{'}\text{'}}=A{e}^t(t+2) \end{gathered}$$

$$\begin{gathered} y_p^{\text{'}\text{'}\text{'}}=A{e}^t(t+3) \\ {y}_p^{\text{'}\text{'}\text{'}}+y_p^{\text{'}}-2y_p=e^t-1 \end{gathered}$$

Substituting value & comparing we get: 

$$\begin{gathered} A{e}^t(t+3+t+1-2t)-2B=e^t-1 \\ 4A=1,-2B=-1 \\ A=\frac{1}{4},B=\frac{1}{2} \end{gathered}$$

So $y=C_1{e}^t+C_2{e}^{\frac{-t}{2}}\cos \left(\frac{\sqrt{7}}{2}t\right)+C_3{e}^{\frac{-t}{2}}\sin \left(\frac{\sqrt{7}}{2}t\right)+\frac{t{e}^t}{4}+\frac{1}{2}$

Step 5: Compute value x

Now we need to substitute value of $y\text{'}\text{'}$in $x=y^{\text{'}\text{'}}+t+1$.

$$\begin{gathered} y^{\text{'}}=C_1{e}^{\text{'}}-\frac{C_2}{2}{e}^{\frac{-t}{2}}\cos \left(\frac{\sqrt{7}t}{2}\right)-\frac{\sqrt{7}}{2}{C}_2{e}^{\frac{-t}{2}}\sin \left(\frac{\sqrt{7}t}{2}\right)+\frac{\sqrt{7}}{2}{C}_3{e}^{\frac{-t}{2}}\cos \left(\frac{\sqrt{7}t}{2}\right)-\frac{C_3}{2}{e}^{\frac{-t}{2}}\sin \left(\frac{\sqrt{7}t}{2}\right)+\frac{e^t}{4}(t+1) \\ {y}^{\text{'}}=e^{\frac{-t}{2}}\cos \left(\frac{\sqrt{7}t}{2}\right)\left(\frac{\sqrt{7}{C}_3-C_2}{2}\right)+e^{\frac{-t}{2}}\sin \left(\frac{\sqrt{7}t}{2}\right)\left(\frac{-\sqrt{7}{C}_2-C_3}{2}\right)+e^t\left(\frac{t}{4}+C_1+\frac{1}{4}\right) \end{gathered}$$

And

$$\begin{gathered} y^{\text{'}\text{'}}=e^{\frac{-t}{2}}\sin \left(\frac{\sqrt{7}t}{2}\right)\left(\frac{-C_2+C_3\sqrt{7}}{2}\right)\left(\frac{-\sqrt{7}}{2}\right)-\frac{1}{2}{e}^{\frac{t}{2}}\cos \left(\frac{\sqrt{7}t}{2}\right)\left(\frac{\sqrt{7}{C}_3-C_2}{2}\right)-\left(\frac{\sqrt{7}{C}_2+C_3}{2}\right)\left\{\frac{\sqrt{7}}{2}{e}^{\frac{-t}{2}}\cos \left(\frac{\sqrt{7}t}{2}\right)-\frac{1}{2}{e}^{\frac{-t}{2}}\cos \left(\frac{\sqrt{7}t}{2}\right)\right\}+e^t\left(\frac{t}{4}+C_1+\frac{1}{2}\right) \\ {y}^{\text{'}\text{'}}=e^{\frac{-t}{2}}\sin \left(\frac{\sqrt{7}t}{2}\right)\left(\frac{\sqrt{7}{C}_2-3C_3}{4}\right)+e^{\frac{-t}{2}}\cos \left(\frac{\sqrt{7}t}{2}\right)\left(\frac{-3C_2-\sqrt{7}{C}_3}{2}\right)+e^t\left(\frac{t}{4}+C_1+\frac{1}{2}\right) \end{gathered}$$

Substituting values we get:

$x=e^{\frac{-t}{2}}\sin \left(\frac{\sqrt{7}t}{2}\right)\left(\frac{\sqrt{7}{C}_2-3C_3}{4}\right)+e^{\frac{-t}{2}}\cos \left(\frac{\sqrt{7}t}{2}\right)\left(\frac{-3C_2-\sqrt{7}{C}_3}{2}\right)+e^t\left(\frac{t}{4}+C_1+\frac{1}{2}\right)+t+1$

Hence the solution of equation is given by 

$x=e^{\frac{-t}{2}}\sin \left(\frac{\sqrt{7}t}{2}\right)\left(\frac{\sqrt{7}{C}_2-3C_3}{4}\right)+e^{\frac{-t}{2}}\cos \left(\frac{\sqrt{7}t}{2}\right)\left(\frac{-3C_2-\sqrt{7}{C}_3}{2}\right)+e^t\left(\frac{t}{4}+C_1+\frac{1}{2}\right)+t+1$

Therefore the solution is :

$$\begin{gathered} x\left(t\right)=\frac{-3c_1-\sqrt{7}{c}_2}{2}{e}^{-\frac{t}{2}}\cos \frac{\sqrt{7}}{2}t+\frac{\sqrt{7}{c}_1-3c_2}{2}{e}^{-\frac{t}{2}}\sin \frac{\sqrt{7}}{2}t+\left(c_3+\frac{1}{2}\right)e^t+\frac{1}{4}t{e}^t+t+1 \\ y\left(t\right)=c_1{e}^{-\frac{t}{2}}\cos \frac{\sqrt{7}}{2}t+c_2{e}^{-\frac{t}{2}}\sin \frac{\sqrt{7}}{2}t+c_3{e}^t+\frac{1}{4}t{e}^t+1 \end{gathered}$$