Q28E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the form of a particular solution for the differential equation. (Do not evaluate coefficients.) $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{5}{\mathbf{t}}^6{\mathbf{e}}^{3t}$

Step-by-Step Solution

Answer

The particular solution is:

${\mathbf{y}}_p\left(\mathbf{x}\right)\mathbf{=}({\mathbf{A}}_6{\mathbf{t}}^8\mathbf{+}{\mathbf{A}}_5{\mathbf{t}}^7\mathbf{+}{\mathbf{A}}_4{\mathbf{t}}^6\mathbf{+}{\mathbf{A}}_3{\mathbf{t}}^5\mathbf{+}{\mathbf{A}}_2{\mathbf{t}}^4\mathbf{+}{\mathbf{A}}_1{\mathbf{t}}^3\mathbf{+}{\mathbf{A}}_0{\mathbf{t}}^2){\mathbf{e}}^{3t}$

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

The given differential equation is in the form of;

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

 

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 nonnegative 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,

$\mathrm{y}\text{'}\text{'}-6\mathrm{y}\text{'}+9\mathrm{y}=5{\mathrm{t}}^6{\mathrm{e}}^{3\mathrm{t}}......\left(1\right)$

 

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

$\mathrm{y}\text{'}\text{'}-6\mathrm{y}\text{'}+9\mathrm{y}=0$

 

The auxiliary equation for the above equation,

${\mathrm{r}}^2-6\mathrm{r}+9=0$

Step 3: Now find the roots of auxiliary equation

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{r}}^2-6\mathrm{r}+9=0\\ {(\mathrm{r}-3)}^2=0\end{array}$

 

The roots of auxiliary equation are, 

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

The complimentary solution of the given equation is,

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

Step 4: Final 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,

 

$\mathrm{y}\text{'}\text{'}-6\mathrm{y}\text{'}+9\mathrm{y}=5{\mathrm{t}}^6{\mathrm{e}}^{3\mathrm{t}}$

 

Condition satisfied, 

 

M=6, s = 2 if r = 3 is a double root of the associated auxiliary equation.

 

Therefore, the particular solution of equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^2({\mathrm{A}}_6{\mathrm{t}}^6+{\mathrm{A}}_5{\mathrm{t}}^5+{\mathrm{A}}_4{\mathrm{t}}^4+{\mathrm{A}}_3{\mathrm{t}}^3+{\mathrm{A}}_2{\mathrm{t}}^2+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{3\mathrm{t}}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=({\mathrm{A}}_6{\mathrm{t}}^8+{\mathrm{A}}_5{\mathrm{t}}^7+{\mathrm{A}}_4{\mathrm{t}}^6+{\mathrm{A}}_3{\mathrm{t}}^5+{\mathrm{A}}_2{\mathrm{t}}^4+{\mathrm{A}}_1{\mathrm{t}}^3+{\mathrm{A}}_0{\mathrm{t}}^2){\mathrm{e}}^{3\mathrm{t}}\end{array}$