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Expert-verified Found in: Page 186 ### 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 # A nonhomogeneous equation and a particular solution are given. Find a general solution for the equation.${\mathbf{\theta }}{\mathbf{\text{'}}}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{\theta }}{\mathbf{\text{'}}}{\mathbf{-}}{\mathbf{2}}{\mathbf{\theta }}{\mathbf{=}}{\mathbf{1}}{\mathbf{-}}{\mathbf{2}}{\mathbf{t}}{\mathbf{,}}{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}}{{\mathbf{\theta }}}_{p}\left(\mathbf{t}\right){\mathbf{=}}{\mathbf{t}}{\mathbf{-}}{\mathbf{1}}$

The general solution of the given differential equation is $\mathrm{\theta }={\mathrm{c}}_{1}{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-\mathrm{t}}+\mathrm{t}-1$.

See the step by step solution

## Step 1: Write the auxiliary equation of the given differential equation

The differential equation is:

$\mathrm{\theta }\text{'}\text{'}-\mathrm{\theta }\text{'}-2\mathrm{\theta }=1-2\mathrm{t}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\dots \left(1\right)$

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

$\mathrm{\theta }\text{'}\text{'}-\mathrm{\theta }\text{'}-2\mathrm{\theta }=0$

The auxiliary equation for the above equation,

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

## Step 2: Now find the complementary solution of the given equation is

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^{2}-\mathrm{m}-2=0\\ {\mathrm{m}}^{2}-2\mathrm{m}+\mathrm{m}-2=0\\ \mathrm{m}\left(\mathrm{m}-2\right)+1\left(\mathrm{m}-2\right)=0\\ \left(\mathrm{m}-2\right)\left(\mathrm{m}+1\right)=0\end{array}$

The roots of the auxiliary equation are,

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

The complementary solution of the given equation is,

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

## Step 3: Use the given particular solution to find a general solution for the equation.

The given particular solution,

${\mathrm{\theta }}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}-1$

Therefore, the general solution is,

$\begin{array}{c}\mathrm{\theta }={\mathrm{\theta }}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{\theta }}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{\theta }={\mathrm{c}}_{1}{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{c}}_{2}{\mathrm{e}}^{-\mathrm{t}}+\mathrm{t}-1\end{array}$ ### Want to see more solutions like these? 