Q34E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 34, use the method of undetermined coefficients to find a particular solution to the given higher-order equation. $\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}{\mathbf{e}}^{-t}$

Step-by-Step Solution

Answer

The particular solution is ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\frac{1}{4}{\mathrm{e}}^{-\mathrm{t}}$

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

The given differential equation is:

 

$2\mathrm{y}\text{'}\text{'}\text{'}+3\mathrm{y}\text{'}\text{'}+\mathrm{y}\text{'}-4\mathrm{y}={\mathrm{e}}^{-\mathrm{t}}\dots \left(1\right)$

 

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

 

$2\mathrm{y}\text{'}\text{'}\text{'}+3\mathrm{y}\text{'}\text{'}+\mathrm{y}\text{'}-4\mathrm{y}=0$

 

The auxiliary equation for the above equation,

 

$2{\mathrm{m}}^3+3{\mathrm{m}}^2+\mathrm{m}-4=0$

 

Check for m = -1,

$\begin{array}{c}2{(-1)}^3+3{(-1)}^2+(-1)-4=0\\ -2+3-1-4=0\\ -4\ne 0\end{array}$

Therefore, m = -1 is not the root of the auxiliary equation.

Step 2: Use the method of undetermined coefficients to find a particular solution to the given differential equation.

Consider the particular solution is,

 

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\mathrm{t}}\dots \left(2\right)$

 

Take the first, second, and third derivative of the above equation,

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=-{\mathrm{Ae}}^{-\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\text{'}\left(\mathrm{t}\right)=-{\mathrm{Ae}}^{-\mathrm{t}}\end{array}$

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

$\begin{array}{c}2\mathrm{y}\text{'}\text{'}\text{'}+3\mathrm{y}\text{'}\text{'}+\mathrm{y}\text{'}-4\mathrm{y}={\mathrm{e}}^{-\mathrm{t}}\\ -2{\mathrm{Ae}}^{-\mathrm{t}}+3{\mathrm{Ae}}^{-\mathrm{t}}+(-{\mathrm{Ae}}^{-\mathrm{t}})-4{\mathrm{Ae}}^{-\mathrm{t}}={\mathrm{e}}^{-\mathrm{t}}\\ -4{\mathrm{Ae}}^{-\mathrm{t}}={\mathrm{e}}^{-\mathrm{t}}\end{array}$ 

 

Comparing the coefficients of the above equation;

  $\begin{array}{c}-4\mathrm{A}=1\\ \mathrm{A}=-\frac{1}{4}\end{array}$

Step 3: Conclusion

Substitute value A in the equation (2),

 

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{Ae}}^{-\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\frac{1}{4}{\mathrm{e}}^{-\mathrm{t}}\end{array}$

 

Therefore, the particular solution of the equation (1),

 

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)=-\frac{1}{4}{\mathrm{e}}^{-\mathrm{t}}$