Q30E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Determine the form of a particular solution for the differential equation. (Do not evaluate coefficients.) $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{7}{\mathbf{e}}^t\mathbf{cost}$

Step-by-Step Solution

Answer

The particular solution is  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{A}}_0{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}+{\mathrm{B}}_0{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}$.

  

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, $\mathbf{\beta }\ne \mathbf{0}$ 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 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,

$\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+\mathrm{y}=7{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}.....\left(1\right)$

 

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

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

 

The auxiliary equation for the above equation,

${\mathrm{m}}^2-2\mathrm{m}+1=0$

Step 3: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2-2\mathrm{m}+1=0\\ {(\mathrm{m}-1)}^2=0\end{array}$

The roots of the auxiliary equation are, 

 

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

 

The complementary solution of the given equation is,

 

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

 

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,

 

$\mathrm{y}\text{'}\text{'}-2\mathrm{y}\text{'}+\mathrm{y}=7{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}$

 

We have,

 $\mathrm{m}=0,\mathrm{\alpha }=1,\mathrm{\beta }=1$

 

And  

 

$\mathrm{\alpha }+\mathrm{i\beta }=1+\mathrm{i}\ne {\mathrm{m}}_1={\mathrm{m}}_2$

 

Therefore, we get

 

s = 0 if  $\mathbf{\alpha }\mathbf{+}\mathbf{i\beta }$ is not a root of the associated auxiliary equation

 

The particular solution to the differential equation for m = 0,

$\begin{array}{l}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^{\mathrm{s}}[({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}]\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^0[\left({\mathrm{A}}_0\right){\mathrm{e}}^{\left(1\right)\mathrm{t}}\cos \left(1\right)\mathrm{t}+{\mathrm{t}}^0\left({\mathrm{B}}_0\right){\mathrm{e}}^{\left(1\right)\mathrm{t}}\sin \left(1\right)\mathrm{t}]\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{A}}_0{\mathrm{e}}^{\mathrm{t}}\mathrm{cost}+{\mathrm{B}}_0{\mathrm{e}}^{\mathrm{t}}\mathrm{sint}\end{array}$