Q5E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

A nonhomogeneous equation and a particular solution are given. Find a general solution for the equation.$\mathbf{\theta }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{\theta }\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{\theta }\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{2}\mathbf{t}\mathbf{,}{\mathbf{\theta }}_p\left(\mathbf{t}\right)\mathbf{=}\mathbf{t}\mathbf{-}\mathbf{1}$

Step-by-Step Solution

Answer

The general solution of the given differential equation is $\mathrm{\theta }={\mathrm{c}}_1{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}+\mathrm{t}-1$.

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

The differential equation is:

 

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

 

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,

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

 

The roots of the auxiliary equation are, 

${\mathrm{m}}_1=2,{\mathrm{m}}_2=-1$

 

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,

 $\begin{array}{c}\mathrm{\theta }={\mathrm{\theta }}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{\theta }}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{\theta }={\mathrm{c}}_1{\mathrm{e}}^{2\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}+\mathrm{t}-1\end{array}$