Q21E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the differential equation. $\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{\theta }\right)\mathbf{+}\mathbf{2}\mathbf{y}\left(\mathbf{\theta }\right)\mathbf{=}{\mathbf{e}}^{-\theta }\mathbf{cos\theta }$

Step-by-Step Solution

Answer

The general solution to the differential equation is:  $\mathbf{y}\mathbf{=}{\mathbf{e}}^{-\theta }[{\mathbf{c}}_1\mathbf{cos\theta }\mathbf{+}{\mathbf{c}}_2\mathbf{sin\theta }]\mathbf{+}\frac{1}{2}{\mathbf{\theta e}}^{-\theta }\mathbf{sin\theta }$

Step 1: Write the auxiliary equation of the given differential equation.

The differential equation is,

 

$\mathrm{y}\text{'}\text{'}\left(\mathrm{\theta }\right)+2\mathrm{y}\text{'}\left(\mathrm{\theta }\right)+2\mathrm{y}\left(\mathrm{\theta }\right)={\mathrm{e}}^{-\mathrm{\theta }}\mathrm{cos\theta } \dots \left(1\right)$

 

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

 

$\mathrm{y}\text{'}\text{'}\left(\mathrm{\theta }\right)+2\mathrm{y}\text{'}\left(\mathrm{\theta }\right)+2\mathrm{y}\left(\mathrm{\theta }\right)=0$

 

The auxiliary equation for the above equation,

 

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

Step 2: Now find the complementary solution of the given equation.

Solve the auxiliary equation,

$\begin{array}{c}{\mathrm{m}}^2+2\mathrm{m}+2=0\\ \mathrm{m}=\frac{-2\pm \sqrt{4-8}}{2}\\ \mathrm{m}=\frac{-2\pm \sqrt{-4}}{2}\\ \mathrm{m}=-1\pm \mathrm{i}\end{array}$

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,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{e}}^{-\mathrm{\theta }}[{\mathrm{c}}_1\mathrm{cos\theta }+{\mathrm{c}}_2\mathrm{sin\theta }]$

Step 3: Use the method of undetermined coefficients.

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

 $\mathrm{ay}\text{'}\text{'}+\mathrm{by}\text{'}+\mathrm{cy}=\left\{\begin{array}{l}{\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{cos\beta t}\\ \mathrm{or}\\ {\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{\alpha t}}\mathrm{sin\beta t}\end{array}\right\}......\left(2\right)$

For $\mathrm{\beta }\ne 0$, 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 4: Find the particular solution to find a general solution for the equation.

Comparing equations (1) and (2), we get;

 

M=0 and $\mathrm{\alpha }=-1;\mathrm{\beta }=1$

 

Therefore,$\mathrm{\alpha }+\mathrm{i\beta }=-1+\mathrm{i}$ is a root of the associated auxiliary equation so here s =1.

Assume, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{s}}({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{\alpha x}}\mathrm{cos\beta x}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{\alpha x}}\mathrm{sin\beta x}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)={\mathrm{\theta }}^1\left({\mathrm{A}}_0\right){\mathrm{e}}^{(-1)\mathrm{\theta }}\cos \left(1\right)\mathrm{\theta }+{\mathrm{\theta }}^1\left({\mathrm{B}}_0\right){\mathrm{e}}^{(-1)\mathrm{\theta }}\sin \left(1\right)\mathrm{\theta }\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)={\mathrm{\theta e}}^{-\mathrm{\theta }}({\mathrm{A}}_0\mathrm{cos\theta }+{\mathrm{B}}_0\mathrm{sin\theta })\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)={\mathrm{\theta e}}^{-\mathrm{\theta }}(\mathrm{Acos\theta }+\mathrm{Bsin\theta })......\left(3\right)\end{array}$

Now find the first and second derivatives of the above equation,

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{\theta }\right)={\mathrm{e}}^{-\mathrm{\theta }}[\mathrm{A}(-\mathrm{\theta sin\theta }+\mathrm{cos\theta })+\mathrm{B}(\mathrm{\theta cos\theta }+\mathrm{sin\theta })]-{\mathrm{\theta e}}^{-\mathrm{\theta }}(\mathrm{Acos\theta }+\mathrm{Bsin\theta })\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{\theta }\right)={\mathrm{e}}^{-\mathrm{\theta }}[(-\mathrm{A}-\mathrm{B})\mathrm{\theta sin\theta }+\mathrm{Acos\theta }+(\mathrm{B}-\mathrm{A})\mathrm{\theta cos\theta }+\mathrm{Bsin\theta }]\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{\theta }\right)={\mathrm{e}}^{-\mathrm{\theta }}[\left(2\mathrm{A}\right)\mathrm{\theta sin\theta }+(2\mathrm{B}-2\mathrm{A})\mathrm{cos\theta }+(-2\mathrm{B})\mathrm{\theta cos\theta }+(-2\mathrm{A}-2\mathrm{B})\mathrm{sin\theta }]\end{array}$

Substitute the value of  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{\theta }\right)$and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{\theta }\right)$ the equation (1),

$\begin{array}{r}\Rightarrow \mathrm{y}\text{'}\text{'}\left(\mathrm{\theta }\right)+2\mathrm{y}\text{'}\left(\mathrm{\theta }\right)+2\mathrm{y}\left(\mathrm{\theta }\right)={\mathrm{e}}^{-\mathrm{\theta }}\mathrm{cos\theta }\\ \Rightarrow {\mathrm{e}}^{-\mathrm{\theta }}[\left(2\mathrm{A}\right)\mathrm{\theta sin\theta }+(2\mathrm{B}-2\mathrm{A})\mathrm{cos\theta }+(-2\mathrm{B})\mathrm{\theta cos\theta }+(-2\mathrm{A}-2\mathrm{B})\mathrm{sin\theta }]\\ +2{\mathrm{e}}^{-\mathrm{\theta }}[(-\mathrm{A}-\mathrm{B})\mathrm{\theta sin\theta }+\mathrm{Acos\theta }+(\mathrm{B}-\mathrm{A})\mathrm{\theta cos\theta }+\mathrm{Bsin\theta }]\\ +2{\mathrm{\theta e}}^{-\mathrm{\theta }}(\mathrm{Acos\theta }+\mathrm{Bsin\theta })={\mathrm{e}}^{-\mathrm{\theta }}\mathrm{cos\theta }\\ \Rightarrow {\mathrm{e}}^{-\mathrm{\theta }}[\left(2\mathrm{B}\right)\mathrm{cos\theta }+(-2\mathrm{A})\mathrm{sin\theta }]={\mathrm{e}}^{-\mathrm{\theta }}\mathrm{cos\theta }\end{array}$

Comparing all coefficients of the above equation,

  $\begin{array}{rcl}-2\mathrm{A}& =& 0\Rightarrow \mathrm{A}=0\\ 2\mathrm{B}& =& 1\Rightarrow \mathrm{B}=\frac{1}{2}\end{array}$

Substitute the value of A and B in the equation (3),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)={\mathrm{\theta e}}^{-\mathrm{\theta }}(\mathrm{Acos\theta }+\mathrm{Bsin\theta })\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)={\mathrm{\theta e}}^{-\mathrm{\theta }}(\left(0\right)\mathrm{cos\theta }+\left(\frac{1}{2}\right)\mathrm{sin\theta })\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=\frac{1}{2}{\mathrm{\theta e}}^{-\mathrm{\theta }}\mathrm{sin\theta }\end{array}$

Therefore, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right)=\frac{1}{2}{\mathrm{\theta e}}^{-\mathrm{\theta }}\mathrm{sin\theta }$

Step 5: Conclusion.

Therefore, the general solution is,

$$\begin{gathered} \mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{\theta }\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{\theta }\right) \\ \mathrm{y}={\mathrm{e}}^{-\mathrm{\theta }}[{\mathrm{c}}_1\mathrm{cos\theta }+{\mathrm{c}}_2\mathrm{sin\theta }]+\frac{1}{2}{\mathrm{\theta e}}^{-\mathrm{\theta }}\mathrm{sin\theta } \end{gathered}$$