Q26E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a particular solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{4}{\mathbf{te}}^{-t}\mathbf{cost}$

Step-by-Step Solution

Answer

The particular solution is ${\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{-\mathrm{t}}[\mathrm{tcost}+{\mathrm{t}}^2\mathrm{sint}]$.

Step 1: Firstly, write the auxiliary equation of the above differential equation.

The differential equation is:

 

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+2\mathrm{y}=4{\mathrm{te}}^{-\mathrm{t}}\mathrm{cost}......\left(1\right)$

 

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

 

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

 

The auxiliary equation for the above equation,

${\mathrm{m}}^2+2\mathrm{m}+2=0$

Step 2: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

 $\begin{array}{c}{\mathrm{m}}^2+2\mathrm{m}+2=0\\ \mathrm{m}=\frac{-2\pm \sqrt{2^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}\\ \mathrm{m}=\frac{-2\pm \sqrt{-4}}{2}\\ \mathrm{m}=-1\pm \mathrm{i}\end{array}$

 

The roots of the auxiliary equation are, 

 

${\mathbf{m}}_1\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{i}\mathbf{,}\&{\mathbf{m}}_2\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{i}$

 

The complementary solution of the given equation is,

 ${\mathrm{y}}_{\mathrm{c}}={\mathrm{e}}^{-\mathrm{x}}({\mathrm{c}}_1\mathrm{cosx}+{\mathrm{c}}_2\mathrm{sinx})$

Step 3: Use the method of undetermined coefficients to find a particular solution to the differential equation

Assume, the particular solution of equation (1),

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=\mathrm{t}[{\mathrm{e}}^{-\mathrm{t}}(\mathrm{At}+\mathrm{B})\mathrm{cost}+{\mathrm{e}}^{-\mathrm{t}}(\mathrm{Ct}+\mathrm{D})\mathrm{sint}]\\ {\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{-\mathrm{t}}[{\mathrm{At}}^2\mathrm{cost}+\mathrm{Btcost}+{\mathrm{Ct}}^2\mathrm{sint}+\mathrm{Dtsint}]......\left(2\right)\end{array}$

 

Now find the derivative of the above equation,

  $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=-{\mathrm{e}}^{-\mathrm{t}}[{\mathrm{At}}^2\mathrm{cost}+\mathrm{Btcost}+{\mathrm{Ct}}^2\mathrm{sint}+\mathrm{Dtsint}]\\ +{\mathrm{e}}^{-\mathrm{t}}[2\mathrm{Atcost}-{\mathrm{At}}^2\mathrm{sint}+\mathrm{Bcost}-\mathrm{Btsint}+2\mathrm{Ctsint}+{\mathrm{Ct}}^2\mathrm{cost}+\mathrm{Dsint}+\mathrm{Dtcost}]\\ \\ {\mathrm{y}}_{\mathrm{p}}\text{'}={\mathrm{e}}^{-\mathrm{t}}[(-\mathrm{A}+\mathrm{C}){\mathrm{t}}^2\mathrm{cost}+(-\mathrm{B}+\mathrm{D}+2\mathrm{A})\mathrm{tcost}+(-\mathrm{C}-\mathrm{A}){\mathrm{t}}^2\mathrm{sint}+(-\mathrm{D}-\mathrm{B}+2\mathrm{C})\mathrm{tsint}+\mathrm{Bcost}+\mathrm{Dsint}]\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=-{\mathrm{e}}^{-\mathrm{t}}\left[\begin{array}{l}(-\mathrm{A}+\mathrm{C}){\mathrm{t}}^2\mathrm{cost}+(-\mathrm{B}+\mathrm{D}+2\mathrm{A})\mathrm{tcost}+(-\mathrm{C}-\mathrm{A}){\mathrm{t}}^2\mathrm{sint}\\ +(-\mathrm{D}-\mathrm{B}+2\mathrm{C})\mathrm{tsint}+\mathrm{Bcost}+\mathrm{Dsint}\end{array}\right]\\ +{\mathrm{e}}^{-\mathrm{t}}\left[\begin{array}{l}2(-\mathrm{A}+\mathrm{C})\mathrm{tcost}-(-\mathrm{A}+\mathrm{C}){\mathrm{t}}^2\mathrm{sint}+(-\mathrm{B}+\mathrm{D}+2\mathrm{A})\mathrm{cost}-(-\mathrm{B}+\mathrm{D}+2\mathrm{A})\mathrm{tsint}\\ +2(-\mathrm{C}-\mathrm{A})\mathrm{tsint}+(-\mathrm{C}-\mathrm{A}){\mathrm{t}}^2\mathrm{cost}+(-\mathrm{D}-\mathrm{B}+2\mathrm{C})\mathrm{sint}+(-\mathrm{D}-\mathrm{B}+2\mathrm{C})\mathrm{tcost}-\mathrm{Bsint}+\mathrm{Dcost}\end{array}\right]\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}={\mathrm{e}}^{-\mathrm{t}}\left[\begin{array}{l}(-2\mathrm{C}){\mathrm{t}}^2\mathrm{cost}+(-2\mathrm{D}-4\mathrm{A}+4\mathrm{C})\mathrm{tcost}+\left(2\mathrm{A}\right){\mathrm{t}}^2\mathrm{sint}\\ +(2\mathrm{B}-4\mathrm{C}-4\mathrm{A})\mathrm{tsint}+(-2\mathrm{B}+2\mathrm{D}+2\mathrm{A})\mathrm{cost}+(-2\mathrm{D}-2\mathrm{B}+2\mathrm{C})\mathrm{sint}\end{array}\right]\end{array}$

From the equation (1), Substitute the value of $y\text{'}{\text{'}}_{p, } y{\text{'}}_p$ and $y_p$ in the equation (1),

$\begin{array}{l}\Rightarrow {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+2{\mathrm{y}}_{\mathrm{p}}\text{'}+2{\mathrm{y}}_{\mathrm{p}}=4{\mathrm{te}}^{-\mathrm{t}}\mathrm{cost}\\ \Rightarrow {\mathrm{e}}^{-\mathrm{t}}\left[\begin{array}{l}(-2\mathrm{C}){\mathrm{t}}^2\mathrm{cost}+(-2\mathrm{D}-4\mathrm{A}+4\mathrm{C})\mathrm{tcost}+\left(2\mathrm{A}\right){\mathrm{t}}^2\mathrm{sint}\\ +(2\mathrm{B}-4\mathrm{C}-4\mathrm{A})\mathrm{tsint}+(-2\mathrm{B}+2\mathrm{D}+2\mathrm{A})\mathrm{cost}+(-2\mathrm{D}-2\mathrm{B}+2\mathrm{C})\mathrm{sint}\end{array}\right]\\ +2{\mathrm{e}}^{-\mathrm{t}}[(-\mathrm{A}+\mathrm{C}){\mathrm{t}}^2\mathrm{cost}+(-\mathrm{B}+\mathrm{D}+2\mathrm{A})\mathrm{tcost}+(-\mathrm{C}-\mathrm{A}){\mathrm{t}}^2\mathrm{sint}+(-\mathrm{D}-\mathrm{B}+2\mathrm{C})\mathrm{tsint}+\mathrm{Bcost}+\mathrm{Dsint}]\\ +2{\mathrm{e}}^{-\mathrm{t}}[{\mathrm{At}}^2\mathrm{cost}+\mathrm{Btcost}+{\mathrm{Ct}}^2\mathrm{sint}+\mathrm{Dtsint}]=4{\mathrm{te}}^{-\mathrm{t}}\mathrm{cost}\\ \Rightarrow {\mathrm{e}}^{-\mathrm{t}}[\left(4\mathrm{C}\right)\mathrm{tcost}-\left(4\mathrm{A}\right)\mathrm{tsint}+(2\mathrm{D}+2\mathrm{A})\mathrm{cost}+(-2\mathrm{B}+2\mathrm{C})\mathrm{sint}]=4{\mathrm{te}}^{-\mathrm{t}}\mathrm{cost}\end{array}$

Step 4: Final conclusion.

Comparing all coefficients of the above equation;

 $\begin{array}{c}4\mathrm{C}=4\Rightarrow \mathrm{C}=1\\ 4\mathrm{A}=0\Rightarrow \mathrm{A}=0\\ 2\mathrm{D}+2\mathrm{A}=0.....\left(3\right)\\ -2\mathrm{B}+2\mathrm{C}=0......\left(4\right)\end{array}$

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

$\begin{array}{c}2\mathrm{D}+2\left(0\right)=0\\ \mathrm{D}=0\end{array}$

 

 

Substitute the value of C in the equation (4),

 $\begin{array}{c}-2\mathrm{B}+2\left(1\right)=0\\ \mathrm{B}=1\end{array}$

 

 

Therefore, the particular solution of equation (1),

 

$\begin{array}{l}{\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{-\mathrm{t}}[{\mathrm{At}}^2\mathrm{cost}+\mathrm{Btcost}+{\mathrm{Ct}}^2\mathrm{sint}+\mathrm{Dtsint}]\\ {\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{-\mathrm{t}}[\mathrm{tcost}+{\mathrm{t}}^2\mathrm{sint}]\end{array}$