Q31E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the form of a particular solution for the differential equation. (Do not evaluate coefficients.) $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{=}\mathbf{8}{\mathbf{t}}^{\mathbf{3}}{\mathbf{e}}^{\mathbf{-}\mathbf{t}}\mathbf{sint}$

Step-by-Step Solution

Answer

The particular solution is given as:${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{t}\left[\left({\mathrm{A}}_3{\mathrm{t}}^3+{\mathrm{A}}_2{\mathrm{t}}^2+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0\right){\mathrm{e}}^{-\mathrm{t}}\mathrm{cost}+\left({\mathrm{B}}_3{\mathrm{t}}^3+{\mathrm{B}}_2{\mathrm{t}}^2+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0\right){\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}\right]$

Step 1: Use the method of undetermined coefficients.

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, $\mathrm{\beta }\ne 0$ then use the form;

 

$({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}$

 

With s = 1 if $\mathrm{\alpha }+\mathrm{i\beta }$  is a root of the associated auxiliary equation, and s = 0 if $\mathrm{\alpha }+\mathrm{i\beta }$ is not a root of the associated auxiliary equation.

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

The differential equation is,

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+2\mathrm{y}=8{\mathrm{t}}^3{\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}\dots \left(1\right)$

 

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

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

 

The auxiliary equation for the above equation,

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

Step 3: Now find the roots of the auxiliary equation

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2+2\mathrm{m}+2=0\\ \mathrm{m}=\frac{-2\pm \sqrt{2^2-4\left(1\right)\left(2\right)}}{2\left(1\right)}\\ \mathrm{m}=\frac{-2\pm \sqrt{-4}}{2}\\ \mathrm{m}=-1\pm \mathrm{i}\end{array}$

The roots of the auxiliary equation are, 

${\mathrm{m}}_1=-1+\mathrm{i},\&{\mathrm{m}}_2=-1-\mathrm{i}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{e}}^{-\mathrm{x}}({\mathrm{c}}_1\mathrm{cosx}+{\mathrm{c}}_2\mathrm{sinx})$

Step 4: Final answer.

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,

 

$\mathrm{y}\text{'}\text{'}+2\mathrm{y}\text{'}+2\mathrm{y}=8{\mathrm{t}}^3{\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}$

 

We have,

 $\mathrm{m}=3,\mathrm{\alpha }=-1,\mathrm{\beta }=1$

 

And  

 $\mathrm{\alpha }+\mathrm{i\beta }=-1+\mathrm{i}={\mathrm{m}}_1$

 

Therefore, we get

 

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

 

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^{\mathrm{s}}[({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}]\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{t}[({\mathrm{A}}_3{\mathrm{t}}^3+{\mathrm{A}}_2{\mathrm{t}}^2+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{-\mathrm{t}}\mathrm{cost}+({\mathrm{B}}_3{\mathrm{t}}^3+{\mathrm{B}}_2{\mathrm{t}}^2+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{-\mathrm{t}}\mathrm{sint}]\end{array}$