Q4E - 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. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{=}\mathbf{1}\mathbf{,}{\mathbf{y}}_p\left(\mathbf{t}\right)\mathbf{=}\mathbf{t}$

Step-by-Step Solution

Answer

$\mathbf{y}\mathbf{=}{\mathbf{c}}_1\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{-t}\mathbf{+}\mathbf{t}$

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

The differential equation is,

 

$\mathrm{y}\text{'}\text{'}+\mathrm{y}\text{'}=1\dots \left(1\right)$

 

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

 

$\mathrm{y}\text{'}\text{'}+\mathrm{y}\text{'}=0$

 

The auxiliary equation for the above equation,

 

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

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

Solve the auxiliary equation,

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

 

The roots of the auxiliary equation are, 

 

${\mathrm{m}}_1=0,{\mathrm{m}}_2=-1$

 

The complementary solution of the given equation is,

 

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1+{\mathrm{c}}_2{\mathrm{e}}^{-\mathrm{t}}$

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

The given particular solution,

 

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=\mathrm{t}$

 

Therefore, the general solution is,

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