Q27E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the form of a particular solution for the differential equation. (Do not evaluate coefficients.) $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{4}{\mathbf{t}}^3\mathbf{sin}\mathbf{3}\mathbf{t}$

Step-by-Step Solution

Answer

The particular solution is:

 ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=({\mathrm{A}}_3{\mathrm{t}}^4+{\mathrm{A}}_2{\mathrm{t}}^3+{\mathrm{A}}_1{\mathrm{t}}^2+{\mathrm{A}}_0\mathrm{t})\cos 3\mathrm{t}+({\mathrm{B}}_3{\mathrm{t}}^4+{\mathrm{B}}_2{\mathrm{t}}^3+{\mathrm{B}}_1{\mathrm{t}}^2+{\mathrm{B}}_0\mathrm{t})\sin 3\mathrm{t}$

Step 1: Use the method of undetermined coefficients.

The given equation is;

 

$\mathrm{y}\text{'}\text{'}+9\mathrm{y}=4{\mathrm{t}}^3\sin 3\mathrm{t}.....\left(1\right)$

 

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

 

$\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}=\left\{\begin{array}{l}{\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}\\ \mathrm{or}\\ {\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}\end{array}\right\}$

 

For, $\mathbf{\beta }\ne \mathbf{0}$ use the form

 

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^{\mathrm{s}}[({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}]$

 

With s = 1 , $\mathbf{\alpha }\mathbf{+}\mathbf{i\beta }$ if is a root of the associated auxiliary equation.

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

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

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{9}\mathbf{y}\mathbf{=}\mathbf{0}$ 

 

The auxiliary equation for the above equation,

${\mathbf{m}}^2\mathbf{+}\mathbf{9}\mathbf{=}\mathbf{0}$ 

Step 3: Now find the roots of an auxiliary equation,

Solve the auxiliary equation,

 

$\begin{array}{c}{\mathrm{m}}^2+9=0\\ \mathrm{m}=\pm 3\mathrm{i}\end{array}$

 

The roots of the auxiliary equation are, 

 

${\mathrm{m}}_1=3\mathrm{i},\&{\mathrm{m}}_2=-3\mathrm{i}$

 

The complementary solution of the given equation is,

 ${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1\cos 3\mathrm{x}+{\mathrm{c}}_2\sin 3\mathrm{x}$

Step 4: Final conclusion.

To find a particular solution to the differential equation 

$\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}=\left\{\begin{array}{l}{\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}\\ \mathrm{or}\\ {\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}\end{array}\right\}$ 

 

Compare with the given differential equation,

 

$\mathrm{y}\text{'}\text{'}+9\mathrm{y}=4{\mathrm{t}}^3\sin 3\mathrm{t}$

 

We have,

$\mathrm{\alpha }=0,\mathrm{\beta }=3$

 

Therefore, one gets

 

S = 1

 

And  

 

$\begin{array}{c}\mathrm{\alpha }+\mathrm{i\beta }=0+3\mathrm{i}\\ ={\mathrm{m}}_1\end{array}$

 

The particular solution to the differential equation for m = 3,

 

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{t}[({\mathrm{A}}_3{\mathrm{t}}^3+{\mathrm{A}}_2{\mathrm{t}}^2+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0)\cos 3\mathrm{t}+({\mathrm{B}}_3{\mathrm{t}}^3+{\mathrm{B}}_2{\mathrm{t}}^2+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0)\sin 3\mathrm{t}]\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=({\mathrm{A}}_3{\mathrm{t}}^4+{\mathrm{A}}_2{\mathrm{t}}^3+{\mathrm{A}}_1{\mathrm{t}}^2+{\mathrm{A}}_0\mathrm{t})\cos 3\mathrm{t}+({\mathrm{B}}_3{\mathrm{t}}^4+{\mathrm{B}}_2{\mathrm{t}}^3+{\mathrm{B}}_1{\mathrm{t}}^2+{\mathrm{B}}_0\mathrm{t})\sin 3\mathrm{t}\end{array}$