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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.${\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\mathbf{\theta }\right){\mathbf{-}}{\mathbf{7}}{\mathbf{y}}{\mathbf{\text{'}}}\left(\mathbf{\theta }\right){\mathbf{=}}{{\mathbf{\theta }}}^{2}$

The particular solution of the differential equation is ${\mathbf{y}}_{p}\mathbf{=}\mathbf{-}\frac{1}{21}{\mathbf{\theta }}^{3}\mathbf{-}\frac{1}{49}{\mathbf{\theta }}^{2}\mathbf{-}\frac{2}{343}\mathbf{\theta }$

See the step by step solution

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

The differential equation is $\mathrm{y}\text{'}\text{'}\left(\mathrm{\theta }\right)-7\mathrm{y}\text{'}\left(\mathrm{\theta }\right)={\mathrm{\theta }}^{2}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

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

$\mathrm{y}\text{'}\text{'}\left(\mathrm{\theta }\right)-7\mathrm{y}\text{'}\left(\mathrm{\theta }\right)=0$

The auxiliary equation for the above equation,

${\mathrm{m}}^{2}-7\mathrm{m}=0$

Step 2: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^{2}-7\mathrm{m}=0\\ \mathrm{m}\left(\mathrm{m}-7\right)=0\end{array}$

The roots of the auxiliary equation are,

${\mathrm{m}}_{1}=0,\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}&\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=7$

The complementary solution of the given equation is,

$\begin{array}{l}{\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_{1}{\mathrm{e}}^{\left(0\right)\mathrm{\theta }}+{\mathrm{c}}_{2}{\mathrm{e}}^{7\mathrm{\theta }}\\ {\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_{1}+{\mathrm{c}}_{2}{\mathrm{e}}^{7\mathrm{\theta }}\end{array}$

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),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}=\mathrm{\theta }\left({\mathrm{A}}_{2}{\mathrm{\theta }}^{2}+{\mathrm{A}}_{1}\mathrm{\theta }+{\mathrm{A}}_{0}\right){\mathrm{e}}^{\left(0\right)\mathrm{\theta }}\\ {\mathrm{y}}_{\mathrm{p}}={\mathrm{A}}_{2}{\mathrm{\theta }}^{3}+{\mathrm{A}}_{1}{\mathrm{\theta }}^{2}+{\mathrm{A}}_{0}\mathrm{\theta }\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(2\right)\end{array}$

Now find the derivative of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}=3{\mathrm{A}}_{2}{\mathrm{\theta }}^{2}+2{\mathrm{A}}_{1}\mathrm{\theta }+{\mathrm{A}}_{0}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}=6{\mathrm{A}}_{2}\mathrm{\theta }+2{\mathrm{A}}_{1}\end{array}$

From the equation (1), substitute the value of ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$,

We get,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{\theta }\right)-7{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{\theta }\right)={\mathrm{\theta }}^{2}\\ 6{\mathrm{A}}_{2}\mathrm{\theta }+2{\mathrm{A}}_{1}-7\left(3{\mathrm{A}}_{2}{\mathrm{\theta }}^{2}+2{\mathrm{A}}_{1}\mathrm{\theta }+{\mathrm{A}}_{0}\right)={\mathrm{\theta }}^{2}\\ 6{\mathrm{A}}_{2}\mathrm{\theta }+2{\mathrm{A}}_{1}-21{\mathrm{A}}_{2}{\mathrm{\theta }}^{2}-14{\mathrm{A}}_{1}\mathrm{\theta }-7{\mathrm{A}}_{0}={\mathrm{\theta }}^{2}\\ -21{\mathrm{A}}_{2}{\mathrm{\theta }}^{2}+\left(-14{\mathrm{A}}_{1}+6{\mathrm{A}}_{2}\right)\mathrm{\theta }+\left(-7{\mathrm{A}}_{0}+2{\mathrm{A}}_{1}\right)={\mathrm{\theta }}^{2}\end{array}$

Step 4: Final conclusion.

Comparing all coefficients of the above equation;

Substitute the value of ${\mathrm{A}}_{2}$ in the equation (3),

role="math" localid="1654872907123" $\begin{array}{c}-14{\mathrm{A}}_{1}+6\left(\frac{-1}{21}\right)=0\\ -14{\mathrm{A}}_{1}=\frac{2}{7}\\ {\mathrm{A}}_{1}=\frac{-1}{49}\end{array}$

Substitute the value of ${A}_{1}$in the equation (4),

role="math" localid="1654872999674" $\begin{array}{c}-7{\mathrm{A}}_{0}+2\left(\frac{-1}{49}\right)=0\\ {\mathrm{A}}_{0}=-\frac{2}{343}\end{array}$

Substitute the value of ${\mathbf{A}}_{0}\mathbf{,}\text{\hspace{0.17em}\hspace{0.17em}}{\mathbf{A}}_{1}$ and ${\mathbf{A}}_{2}$ in the equation (2),

$\begin{array}{c}{\mathbf{y}}_{p}\mathbf{=}{\mathbf{A}}_{2}{\mathbf{\theta }}^{3}\mathbf{+}{\mathbf{A}}_{1}{\mathbf{\theta }}^{2}\mathbf{+}{\mathbf{A}}_{0}\mathbf{\theta }\\ {\mathbf{y}}_{p}\mathbf{=}\mathbf{-}\frac{1}{21}{\mathbf{\theta }}^{3}\mathbf{-}\frac{1}{49}{\mathbf{\theta }}^{2}\mathbf{-}\frac{2}{343}\mathbf{\theta }\end{array}$

Therefore, the particular solution of equation (1),

${\mathbf{y}}_{p}\mathbf{=}\mathbf{-}\frac{1}{21}{\mathbf{\theta }}^{3}\mathbf{-}\frac{1}{49}{\mathbf{\theta }}^{2}\mathbf{-}\frac{2}{343}\mathbf{\theta }$