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Q16E

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

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Find a particular solution to the differential equation.${\mathbf{\theta }}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\mathbf{t}\right){\mathbf{-}}{\mathbf{\theta }}\left(\mathbf{t}\right){\mathbf{=}}{\mathbf{tsint}}$

The particular solution of the differential equation is ${\mathbf{\theta }}_{p}\mathbf{=}\mathbf{-}\frac{1}{2}\mathbf{tsin}\left(\mathbf{t}\right)\mathbf{-}\frac{1}{2}\mathbf{cos}\left(\mathbf{t}\right)$.

See the step by step solution

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

The differential equation;

$\mathrm{\theta }\text{'}\text{'}\left(\mathrm{t}\right)-\mathrm{\theta }\left(\mathrm{t}\right)=\mathrm{tsint}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

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

$\mathrm{\theta }\text{'}\text{'}\left(\mathrm{t}\right)-\mathrm{\theta }\left(\mathrm{t}\right)=0$

The auxiliary equation for the above equation,

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

Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation:

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

The roots of the auxiliary equation are:

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

The complementary solution of the given equation is:

${\mathrm{\theta }}_{\mathrm{c}}={\mathrm{c}}_{1}{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-\mathrm{x}}$

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),

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

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{\theta }}_{\mathrm{p}}\text{'}=\left(\mathrm{At}+\mathrm{B}\right)\mathrm{cost}+\mathrm{Asint}+\left(\mathrm{Ct}+\mathrm{D}\right)\left(-\mathrm{sint}\right)+\mathrm{Ccost}\\ {\mathrm{\theta }}_{\mathrm{p}}\text{'}\text{'}=\left(\mathrm{At}+\mathrm{B}\right)\left(-\mathrm{sint}\right)+\mathrm{Acost}+\mathrm{Acost}+\left(\mathrm{Ct}+\mathrm{D}\right)\left(-\mathrm{cost}\right)+\mathrm{C}\left(-\mathrm{sint}\right)-\mathrm{Csint}\\ {\mathrm{\theta }}_{\mathrm{p}}\text{'}\text{'}=\left(\mathrm{At}+\mathrm{B}\right)\left(-\mathrm{sint}\right)+2\mathrm{Acost}+\left(\mathrm{Ct}+\mathrm{D}\right)\left(-\mathrm{cost}\right)-2\mathrm{C}\left(\mathrm{sint}\right)\end{array}$

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

$\begin{array}{c}{\mathrm{\theta }}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)-{\mathrm{\theta }}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{tsint}\\ \left(\mathrm{At}+\mathrm{B}\right)\left(-\mathrm{sint}\right)+2\mathrm{Acost}+\left(\mathrm{Ct}+\mathrm{D}\right)\left(-\mathrm{cost}\right)-2\mathrm{Csint}-\left[\left(\mathrm{At}+\mathrm{B}\right)\mathrm{sint}+\left(\mathrm{Ct}+\mathrm{D}\right)\mathrm{cost}\right]=\mathrm{tsint}\\ \left(-2\mathrm{A}\right)\mathrm{tsint}+\left(-2\mathrm{B}-2\mathrm{C}\right)\mathrm{sint}+\left(-2\mathrm{C}\right)\mathrm{tcost}+\left(2\mathrm{A}-2\mathrm{D}\right)\mathrm{cost}=\mathrm{tsint}\end{array}$

Step 4: Final conclusion.

Comparing all coefficients of the above equation;

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

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

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

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

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

Therefore, the particular solution of equation (1),

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