Suggested languages for you:

Americas

Europe

Q26E

Expert-verified
Found in: Page 180

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Find a particular solution to the differential equation.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{=}}{\mathbf{4}}{{\mathbf{te}}}^{-t}{\mathbf{cost}}$

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

See the step by step solution

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

The differential equation is:

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+2\mathrm{y}=4{\mathrm{te}}^{-\mathrm{t}}\mathrm{cost}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(1\right)$

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,

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

The roots of the auxiliary equation are,

${\mathbf{m}}_{1}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{+}\mathbf{i}\mathbf{,}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}&\text{\hspace{0.17em}\hspace{0.17em}}{\mathbf{m}}_{2}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{-}\mathbf{i}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{e}}^{-\mathrm{x}}\left({\mathrm{c}}_{1}\mathrm{cosx}+{\mathrm{c}}_{2}\mathrm{sinx}\right)$

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

Assume, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=\mathrm{t}\left[{\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{At}+\mathrm{B}\right)\mathrm{cost}+{\mathrm{e}}^{-\mathrm{t}}\left(\mathrm{Ct}+\mathrm{D}\right)\mathrm{sint}\right]\\ {\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{-\mathrm{t}}\left[{\mathrm{At}}^{2}\mathrm{cost}+\mathrm{Btcost}+{\mathrm{Ct}}^{2}\mathrm{sint}+\mathrm{Dtsint}\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(2\right)\end{array}$

Now find the derivative of the above equation,

role="math" localid="1654875111330" $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=-{\mathrm{e}}^{-\mathrm{t}}\left[{\mathrm{At}}^{2}\mathrm{cost}+\mathrm{Btcost}+{\mathrm{Ct}}^{2}\mathrm{sint}+\mathrm{Dtsint}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+{\mathrm{e}}^{-\mathrm{t}}\left[2\mathrm{Atcost}-{\mathrm{At}}^{2}\mathrm{sint}+\mathrm{Bcost}-\mathrm{Btsint}+2\mathrm{Ctsint}+{\mathrm{Ct}}^{2}\mathrm{cost}+\mathrm{Dsint}+\mathrm{Dtcost}\right]\\ \\ {\mathrm{y}}_{\mathrm{p}}\text{'}={\mathrm{e}}^{-\mathrm{t}}\left[\left(-\mathrm{A}+\mathrm{C}\right){\mathrm{t}}^{2}\mathrm{cost}+\left(-\mathrm{B}+\mathrm{D}+2\mathrm{A}\right)\mathrm{tcost}+\left(-\mathrm{C}-\mathrm{A}\right){\mathrm{t}}^{2}\mathrm{sint}+\left(-\mathrm{D}-\mathrm{B}+2\mathrm{C}\right)\mathrm{tsint}+\mathrm{Bcost}+\mathrm{Dsint}\right]\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=-{\mathrm{e}}^{-\mathrm{t}}\left[\begin{array}{l}\left(-\mathrm{A}+\mathrm{C}\right){\mathrm{t}}^{2}\mathrm{cost}+\left(-\mathrm{B}+\mathrm{D}+2\mathrm{A}\right)\mathrm{tcost}+\left(-\mathrm{C}-\mathrm{A}\right){\mathrm{t}}^{2}\mathrm{sint}\\ +\left(-\mathrm{D}-\mathrm{B}+2\mathrm{C}\right)\mathrm{tsint}+\mathrm{Bcost}+\mathrm{Dsint}\end{array}\right]\\ +{\mathrm{e}}^{-\mathrm{t}}\left[\begin{array}{l}2\left(-\mathrm{A}+\mathrm{C}\right)\mathrm{tcost}-\left(-\mathrm{A}+\mathrm{C}\right){\mathrm{t}}^{2}\mathrm{sint}+\left(-\mathrm{B}+\mathrm{D}+2\mathrm{A}\right)\mathrm{cost}-\left(-\mathrm{B}+\mathrm{D}+2\mathrm{A}\right)\mathrm{tsint}\\ +2\left(-\mathrm{C}-\mathrm{A}\right)\mathrm{tsint}+\left(-\mathrm{C}-\mathrm{A}\right){\mathrm{t}}^{2}\mathrm{cost}+\left(-\mathrm{D}-\mathrm{B}+2\mathrm{C}\right)\mathrm{sint}+\left(-\mathrm{D}-\mathrm{B}+2\mathrm{C}\right)\mathrm{tcost}-\mathrm{Bsint}+\mathrm{Dcost}\end{array}\right]\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}={\mathrm{e}}^{-\mathrm{t}}\left[\begin{array}{l}\left(-2\mathrm{C}\right){\mathrm{t}}^{2}\mathrm{cost}+\left(-2\mathrm{D}-4\mathrm{A}+4\mathrm{C}\right)\mathrm{tcost}+\left(2\mathrm{A}\right){\mathrm{t}}^{2}\mathrm{sint}\\ +\left(2\mathrm{B}-4\mathrm{C}-4\mathrm{A}\right)\mathrm{tsint}+\left(-2\mathrm{B}+2\mathrm{D}+2\mathrm{A}\right)\mathrm{cost}+\left(-2\mathrm{D}-2\mathrm{B}+2\mathrm{C}\right)\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),

role="math" localid="1654875265020" $\begin{array}{l}⇒{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+2{\mathrm{y}}_{\mathrm{p}}\text{'}+2{\mathrm{y}}_{\mathrm{p}}=4{\mathrm{te}}^{-\mathrm{t}}\mathrm{cost}\\ ⇒{\mathrm{e}}^{-\mathrm{t}}\left[\begin{array}{l}\left(-2\mathrm{C}\right){\mathrm{t}}^{2}\mathrm{cost}+\left(-2\mathrm{D}-4\mathrm{A}+4\mathrm{C}\right)\mathrm{tcost}+\left(2\mathrm{A}\right){\mathrm{t}}^{2}\mathrm{sint}\\ +\left(2\mathrm{B}-4\mathrm{C}-4\mathrm{A}\right)\mathrm{tsint}+\left(-2\mathrm{B}+2\mathrm{D}+2\mathrm{A}\right)\mathrm{cost}+\left(-2\mathrm{D}-2\mathrm{B}+2\mathrm{C}\right)\mathrm{sint}\end{array}\right]\\ +2{\mathrm{e}}^{-\mathrm{t}}\left[\left(-\mathrm{A}+\mathrm{C}\right){\mathrm{t}}^{2}\mathrm{cost}+\left(-\mathrm{B}+\mathrm{D}+2\mathrm{A}\right)\mathrm{tcost}+\left(-\mathrm{C}-\mathrm{A}\right){\mathrm{t}}^{2}\mathrm{sint}+\left(-\mathrm{D}-\mathrm{B}+2\mathrm{C}\right)\mathrm{tsint}+\mathrm{Bcost}+\mathrm{Dsint}\right]\\ +2{\mathrm{e}}^{-\mathrm{t}}\left[{\mathrm{At}}^{2}\mathrm{cost}+\mathrm{Btcost}+{\mathrm{Ct}}^{2}\mathrm{sint}+\mathrm{Dtsint}\right]=4{\mathrm{te}}^{-\mathrm{t}}\mathrm{cost}\\ ⇒{\mathrm{e}}^{-\mathrm{t}}\left[\left(4\mathrm{C}\right)\mathrm{tcost}-\left(4\mathrm{A}\right)\mathrm{tsint}+\left(2\mathrm{D}+2\mathrm{A}\right)\mathrm{cost}+\left(-2\mathrm{B}+2\mathrm{C}\right)\mathrm{sint}\right]=4{\mathrm{te}}^{-\mathrm{t}}\mathrm{cost}\end{array}$

## Step 4: Final conclusion.

Comparing all coefficients of the above equation;

$\begin{array}{c}4\mathrm{C}=4\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{C}=1\\ 4\mathrm{A}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{A}=0\\ 2\mathrm{D}+2\mathrm{A}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.....\left(3\right)\\ -2\mathrm{B}+2\mathrm{C}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(4\right)\end{array}$

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),

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

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.