Q32E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the form of a particular solution for the differential equation. (Do not evaluate coefficients.)  $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{\text{'}}\mathbf{-}\mathbf{12}\mathbf{y}\mathbf{=}\mathbf{2}{\mathbf{t}}^6{\mathbf{e}}^{-3t}$

Step-by-Step Solution

Answer

The particular solution is:

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=({\mathrm{A}}_6{\mathrm{t}}^7+{\mathrm{A}}_5{\mathrm{t}}^6+{\mathrm{A}}_4{\mathrm{t}}^5+{\mathrm{A}}_3{\mathrm{t}}^4+{\mathrm{A}}_2{\mathrm{t}}^3+{\mathrm{A}}_1{\mathrm{t}}^2+{\mathrm{A}}_0\mathrm{t}){\mathrm{e}}^{-3\mathrm{t}}$.

Step 1: Use the method of undetermined coefficients to find a particular solution to the given differential equation

The given differential equation is in the form of,

$\mathrm{ax}\text{'}\text{'}+\mathrm{bx}\text{'}+\mathrm{cx}={\mathrm{e}}^{\mathrm{rt}}$

 

According to the method of undetermined coefficients, to find a particular solution to the differential equation;

 $\mathrm{ay}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{by}\text{'}\left(\mathrm{x}\right)+\mathrm{cy}\left(\mathrm{x}\right)={\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{rt}}$

 

Where m is a non-negative integer, use the form

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^{\mathrm{s}}({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{rt}}$

 

1. s = 0 if r is not a root of the associated auxiliary equation; 
2. s = 1 if r is a simple root of the associated auxiliary equation; 
3.  s = 2 if r is a double root of the associated auxiliary equation.

Step 2: Now, write the auxiliary equation of the above differential equation

The given differential equation is,

 

$\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}-12\mathrm{y}=2{\mathrm{t}}^6{\mathrm{e}}^{-3\mathrm{t}}......\left(1\right)$

 

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

 

$\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}-12\mathrm{y}=0$

 

The auxiliary equation for the above equation,

 ${\mathrm{r}}^2-\mathrm{r}-12=0$

Step 3: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{r}}^2-\mathrm{r}-12=0\\ {\mathrm{r}}^2-4\mathrm{r}+3\mathrm{r}-12=0\\ \mathrm{r}(\mathrm{r}-4)+3(\mathrm{r}-4)=0\\ (\mathrm{r}-4)(\mathrm{r}+3)=0\end{array}$

The roots of the auxiliary equation are, 

${\mathrm{r}}_1=4,\&{\mathrm{r}}_2=-3$

 

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{4\mathrm{t}}+{\mathrm{c}}_2{\mathrm{e}}^{-3\mathrm{t}}$

 

Step 4: Conclusion.

To find a particular solution to the differential equation;

 

$\mathrm{ay}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{by}\text{'}\left(\mathrm{x}\right)+\mathrm{cy}\left(\mathrm{x}\right)={\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{rt}}$

 

Compare with the given differential equation,

 

$\mathrm{y}\text{'}\text{'}-\mathrm{y}\text{'}-12\mathrm{y}=2{\mathrm{t}}^6{\mathrm{e}}^{-3\mathrm{t}}$

 

Condition satisfied, 

 

M=6, s = 1 if r = -3 is a simple root of the associated auxiliary equation;

 

Therefore, the particular solution of the equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^{\mathrm{s}}({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{rt}}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{t}({\mathrm{A}}_6{\mathrm{t}}^6+{\mathrm{A}}_5{\mathrm{t}}^5+{\mathrm{A}}_4{\mathrm{t}}^4+{\mathrm{A}}_3{\mathrm{t}}^3+{\mathrm{A}}_2{\mathrm{t}}^2+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{-3\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=({\mathrm{A}}_6{\mathrm{t}}^7+{\mathrm{A}}_5{\mathrm{t}}^6+{\mathrm{A}}_4{\mathrm{t}}^5+{\mathrm{A}}_3{\mathrm{t}}^4+{\mathrm{A}}_2{\mathrm{t}}^3+{\mathrm{A}}_1{\mathrm{t}}^2+{\mathrm{A}}_0\mathrm{t}){\mathrm{e}}^{-3\mathrm{t}}\end{array}$