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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.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{4}}{\mathbf{y}}{\mathbf{=}}{\mathbf{16}}{\mathbf{tsin}}\left(\mathbf{2}\mathbf{t}\right)$

The particular solution is ${\mathrm{y}}_{\mathrm{p}}=\mathrm{t}\left[\mathrm{sin}\left(2\mathrm{t}\right)-2\mathrm{tcos}\left(2\mathrm{t}\right)\right].$

See the step by step solution

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

The differential equation is $\mathrm{y}\text{'}\text{'}+4\mathrm{y}=16\mathrm{tsin}\left(2\mathrm{t}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

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

$\mathrm{y}\text{'}\text{'}+4\mathrm{y}=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^{2}+4=0$

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

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^{2}+4=0\\ \mathrm{m}=±2\mathrm{i}\end{array}$

The roots of the auxiliary equation are,

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

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_{1}\mathrm{cos}2\mathrm{t}+{\mathrm{c}}_{2}\mathrm{sin}2\mathrm{t}$

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

According to the method of undetermined coefficients, assume, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}=\mathrm{t}\left[\left(\mathrm{At}+\mathrm{B}\right)\mathrm{sin}\left(2\mathrm{t}\right)+\left(\mathrm{Ct}+\mathrm{D}\right)\mathrm{cos}\left(2\mathrm{t}\right)\right]\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(2\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=\left(2{\mathrm{At}}^{2}+2\mathrm{Bt}+2\mathrm{Ct}+\mathrm{D}\right)\mathrm{cos}\left(2\mathrm{t}\right)+\left(2\mathrm{At}-2\mathrm{Dt}-2{\mathrm{Ct}}^{2}+\mathrm{B}\right)\mathrm{sin}\left(2\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=\left(8\mathrm{At}-4{\mathrm{Ct}}^{2}-4\mathrm{tD}+4\mathrm{B}+2\mathrm{C}\right)\mathrm{cos}\left(2\mathrm{t}\right)+\left(-4{\mathrm{At}}^{2}-4\mathrm{D}-8\mathrm{Ct}-4\mathrm{Bt}+2\mathrm{A}\right)\mathrm{sin}\left(2\mathrm{t}\right)\end{array}$

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

role="math" localid="1654868530352" $\begin{array}{l}⇒{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+4{\mathrm{y}}_{\mathrm{p}}=16\mathrm{tsin}\left(2\mathrm{t}\right)\\ ⇒\left(8\mathrm{At}-4{\mathrm{Ct}}^{2}-4\mathrm{tD}+4\mathrm{B}+2\mathrm{C}\right)\mathrm{cos}\left(2\mathrm{t}\right)+\left(-4{\mathrm{At}}^{2}-4\mathrm{D}-8\mathrm{Ct}-4\mathrm{Bt}+2\mathrm{A}\right)\mathrm{sin}\left(2\mathrm{t}\right)\\ +4\left\{\mathrm{t}\left[\left(\mathrm{At}+\mathrm{B}\right)\mathrm{sin}\left(2\mathrm{t}\right)+\left(\mathrm{Ct}+\mathrm{D}\right)\mathrm{cos}\left(2\mathrm{t}\right)\right]\right\}=16\mathrm{tsin}\left(2\mathrm{t}\right)\\ ⇒\mathrm{tsin}\left(2\mathrm{t}\right)\left(-8\mathrm{C}\right)+\mathrm{tcos}\left(2\mathrm{t}\right)\left(8\mathrm{A}\right)+\mathrm{cos}\left(2\mathrm{t}\right)\left(4\mathrm{B}+2\mathrm{C}\right)+\mathrm{sin}\left(2\mathrm{t}\right)\left(2\mathrm{A}-4\mathrm{D}\right)=16\mathrm{tsin}\left(2\mathrm{t}\right)\end{array}$

## Step 4: Final conclusion.

Comparing all coefficients of the above equation;

$\begin{array}{c}8\mathrm{A}=0\text{\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{A}=0\\ -8\mathrm{C}=16\text{\hspace{0.17em}\hspace{0.17em}}⇒\mathrm{C}=-2\\ 4\mathrm{B}+2\mathrm{C}=8\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(3\right)\\ 2\mathrm{A}-4\mathrm{D}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(4\right)\end{array}$

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

$\begin{array}{c}4\mathrm{B}+2\left(-2\right)=8\\ \mathrm{B}=1\end{array}$

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

$\begin{array}{c}2\left(0\right)-4\mathrm{D}=0\\ \mathrm{D}=0\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=\mathrm{t}\left[\left(\mathrm{At}+\mathrm{B}\right)\mathrm{sin}\left(2\mathrm{t}\right)+\left(\mathrm{Ct}+\mathrm{D}\right)\mathrm{cos}\left(2\mathrm{t}\right)\right]\\ {\mathrm{y}}_{\mathrm{p}}=\mathrm{t}\left[\mathrm{sin}\left(2\mathrm{t}\right)-2\mathrm{tcos}\left(2\mathrm{t}\right)\right]\end{array}$ ### Want to see more solutions like these? 