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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.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{4}}{\mathbf{y}}{\mathbf{=}}{\mathbf{111}}{{\mathbf{e}}}^{2t}{\mathbf{cos}}{\mathbf{3}}{\mathbf{t}}$

The particular solution is ${\mathbf{y}}_{p}\mathbf{=}{\mathbf{e}}^{2t}\left(\mathbf{cos}\mathbf{3}\mathbf{t}\mathbf{+}\mathbf{6}\mathbf{sin}\mathbf{3}\mathbf{t}\right)$.

See the step by step solution

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

The differential equation is:

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{111}{\mathbf{e}}^{2t}\mathbf{cos}\mathbf{3}\mathbf{t}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(\mathbf{1}\right)$

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{0}$

The auxiliary equation for the above equation,

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

Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^{2}+2\mathrm{m}+4=0\\ \mathrm{m}=\frac{-2±\sqrt{{2}^{2}-4\left(1\right)\left(4\right)}}{2\left(1\right)}\\ \mathrm{m}=\frac{-2±\sqrt{-12}}{2}\\ \mathrm{m}=-1±\mathrm{i}\sqrt{3}\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_{1}=-1+\mathrm{i}\sqrt{3},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}&\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=-1-\mathrm{i}\sqrt{3}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{e}}^{-\mathrm{x}}\left({\mathrm{c}}_{1}\mathrm{cos}\sqrt{3}\mathrm{x}+{\mathrm{c}}_{2}\mathrm{sin}\sqrt{3}\mathrm{x}\right)$

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),

${\mathbf{y}}_{p}\mathbf{=}{\mathbf{e}}^{2t}\left({\mathbf{A}}_{0}\mathbf{cos}\mathbf{3}\mathbf{t}\mathbf{+}{\mathbf{B}}_{0}\mathbf{sin}\mathbf{3}\mathbf{t}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(\mathbf{2}\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=2{\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{A}}_{0}\mathrm{cos}3\mathrm{t}+{\mathrm{B}}_{0}\mathrm{sin}3\mathrm{t}\right)+{\mathrm{e}}^{2\mathrm{t}}\left(-3{\mathrm{A}}_{0}\mathrm{sin}3\mathrm{t}+3{\mathrm{B}}_{0}\mathrm{cos}3\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=4{\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{A}}_{0}\mathrm{cos}3\mathrm{t}+{\mathrm{B}}_{0}\mathrm{sin}3\mathrm{t}\right)+2{\mathrm{e}}^{2\mathrm{t}}\left(-3{\mathrm{A}}_{0}\mathrm{sin}3\mathrm{t}+3{\mathrm{B}}_{0}\mathrm{cos}3\mathrm{t}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+2{\mathrm{e}}^{2\mathrm{t}}\left(-3{\mathrm{A}}_{0}\mathrm{sin}3\mathrm{t}+3{\mathrm{B}}_{0}\mathrm{cos}3\mathrm{t}\right)+{\mathrm{e}}^{2\mathrm{t}}\left(-9{\mathrm{A}}_{0}\mathrm{cos}3\mathrm{t}-9{\mathrm{B}}_{0}\mathrm{sin}3\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}={\mathrm{e}}^{2\mathrm{t}}\left(-5{\mathrm{A}}_{0}\mathrm{cos}3\mathrm{t}-5{\mathrm{B}}_{0}\mathrm{sin}3\mathrm{t}\right)+{\mathrm{e}}^{2\mathrm{t}}\left(-12{\mathrm{A}}_{0}\mathrm{sin}3\mathrm{t}+12{\mathrm{B}}_{0}\mathrm{cos}3\mathrm{t}\right)\end{array}$

From the equation (1), Substitute the value of ${\mathbf{y}}_{p}\mathbf{\text{'}}\mathbf{\text{'}},\text{\hspace{0.17em}\hspace{0.17em}}{\mathbf{y}}_{p}\mathbf{\text{'}}$and ${\mathbf{y}}_{p}$ in the equation (1),

$\begin{array}{l}⇒{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}+2{\mathrm{y}}_{\mathrm{p}}\text{'}+4{\mathrm{y}}_{\mathrm{p}}=111{\mathrm{e}}^{2\mathrm{t}}\mathrm{cos}3\mathrm{t}\\ ⇒{\mathrm{e}}^{2\mathrm{t}}\left(-5{\mathrm{A}}_{0}\mathrm{cos}3\mathrm{t}-5{\mathrm{B}}_{0}\mathrm{sin}3\mathrm{t}\right)+{\mathrm{e}}^{2\mathrm{t}}\left(-12{\mathrm{A}}_{0}\mathrm{sin}3\mathrm{t}+12{\mathrm{B}}_{0}\mathrm{cos}3\mathrm{t}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+2\left[2{\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{A}}_{0}\mathrm{cos}3\mathrm{t}+{\mathrm{B}}_{0}\mathrm{sin}3\mathrm{t}\right)+{\mathrm{e}}^{2\mathrm{t}}\left(-3{\mathrm{A}}_{0}\mathrm{sin}3\mathrm{t}+3{\mathrm{B}}_{0}\mathrm{cos}3\mathrm{t}\right)\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}+4{\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{A}}_{0}\mathrm{cos}3\mathrm{t}+{\mathrm{B}}_{0}\mathrm{sin}3\mathrm{t}\right)=111{\mathrm{e}}^{2\mathrm{t}}\mathrm{cos}3\mathrm{t}\\ ⇒{\mathrm{e}}^{2\mathrm{t}}\left(3{\mathrm{A}}_{0}\mathrm{cos}3\mathrm{t}+3{\mathrm{B}}_{0}\mathrm{sin}3\mathrm{t}\right)+{\mathrm{e}}^{2\mathrm{t}}\left(-18{\mathrm{A}}_{0}\mathrm{sin}3\mathrm{t}+18{\mathrm{B}}_{0}\mathrm{cos}3\mathrm{t}\right)=111{\mathrm{e}}^{2\mathrm{t}}\mathrm{cos}3\mathrm{t}\\ ⇒{\mathrm{e}}^{2\mathrm{t}}\mathrm{cos}3\mathrm{t}\left(3{\mathrm{A}}_{0}+18{\mathrm{B}}_{0}\right)+{\mathrm{e}}^{2\mathrm{t}}\mathrm{sin}3\mathrm{t}\left(3{\mathrm{B}}_{0}-18{\mathrm{A}}_{0}\right)=111{\mathrm{e}}^{2\mathrm{t}}\mathrm{cos}3\mathrm{t}\end{array}$

Step 4: Final conclusion:

Comparing all coefficients of the above equation;

$\begin{array}{c}3{\mathrm{A}}_{0}+18{\mathrm{B}}_{0}=111\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(3\right)\\ 3{\mathrm{B}}_{0}-18{\mathrm{A}}_{0}=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(4\right)\end{array}$

Solve the above equations,

$\begin{array}{c}\left(3{\mathrm{A}}_{0}+18{\mathrm{B}}_{0}\right)=111×6\\ 18{\mathrm{A}}_{0}+108{\mathrm{B}}_{0}=666\\ -18{\mathrm{A}}_{0}+3{\mathrm{B}}_{0}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}=0\\ \overline{111{\mathrm{B}}_{0}=666\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{B}}_{0}=6\end{array}$

Substitute the value of ${\mathrm{B}}_{0}$ in the equation (3),

$\begin{array}{c}3{\mathrm{A}}_{0}+18\left(6\right)=111\\ 3{\mathrm{A}}_{0}=3\\ {\mathrm{A}}_{0}=1\end{array}$

Therefore, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{2\mathrm{t}}\left({\mathrm{A}}_{0}\mathrm{cos}3\mathrm{t}+{\mathrm{B}}_{0}\mathrm{sin}3\mathrm{t}\right)\\ {\mathrm{y}}_{\mathrm{p}}={\mathrm{e}}^{2\mathrm{t}}\left(\mathrm{cos}3\mathrm{t}+6\mathrm{sin}3\mathrm{t}\right)\end{array}$

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