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Q14E

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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

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Find a particular solution to the differential equation.${\mathbf{2}}{\mathbf{z}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{+}}{\mathbf{z}}{\mathbf{=}}{\mathbf{9}}{{\mathbf{e}}}^{2t}$

The particular solution of the given differential equation is ${\mathrm{z}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{e}}^{2\mathrm{t}}.$

See the step by step solution

Step 1: firstly, write the auxiliary equation of the given differential equation

The given differential equation is,

$2\mathrm{z}\text{'}\text{'}+\mathrm{z}=9{\mathrm{e}}^{2\mathrm{t}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

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

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

The auxiliary equation for the above equation,

$2{\mathrm{m}}^{2}+1=0$

Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

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

The roots of the auxiliary equation are,

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

The complementary solution of the given equation is,

${\mathbf{Z}}_{c}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{c}}_{1}\mathbf{cos}\left(\sqrt{\frac{1}{2}}\mathbf{t}\right)\mathbf{+}{\mathbf{c}}_{2}\mathbf{sin}\left(\sqrt{\frac{1}{2}}\mathbf{t}\right)$

Step 3: Final conclusion, find a particular solution to the differential equation

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

${\mathrm{z}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{Ae}}^{2\mathrm{t}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(2\right)$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{z}}_{\mathrm{p}}\text{'}\left(\mathrm{x}\right)=2{\mathrm{Ae}}^{2\mathrm{t}}\\ {\mathrm{z}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{x}\right)=4{\mathrm{Ae}}^{2\mathrm{t}}\end{array}$

From the equation (1),

$\begin{array}{c}2{\mathrm{z}}_{\mathrm{p}}\text{'}\text{'}+{\mathrm{z}}_{\mathrm{p}}=9{\mathrm{e}}^{2\mathrm{t}}\\ 2\left(4{\mathrm{Ae}}^{2\mathrm{t}}\right)+{\mathrm{Ae}}^{2\mathrm{t}}=9{\mathrm{e}}^{2\mathrm{t}}\\ 9{\mathrm{Ae}}^{2\mathrm{t}}=9{\mathrm{e}}^{2\mathrm{t}}\\ \mathrm{A}=1\end{array}$

Substitute the value of A in the equation (2),

${\mathrm{z}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{e}}^{2\mathrm{t}}$

Therefore, the particular solution of equation (1),

${\mathrm{z}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{e}}^{2\mathrm{t}}$

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