Q19E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Find a general solution to the differential equation.

$\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{-}\mathbf{3}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{+}\mathbf{2}\mathbf{y}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{e}}^x\mathbf{sinx}$

Step-by-Step Solution

Answer

The general solution to the differential equation is $\mathbf{y}\mathbf{=}{\mathbf{c}}_1{\mathbf{e}}^x\mathbf{+}{\mathbf{c}}_2{\mathbf{e}}^{2x}\mathbf{+}\frac{1}{2}{\mathbf{e}}^x(\mathbf{cosx}\mathbf{-}\mathbf{sinx})$.

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

The differential equation is,

 

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-3\mathrm{y}\text{'}\left(\mathrm{x}\right)+2\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx} ......\left(1\right)$

 

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

 

$\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-3\mathrm{y}\text{'}\left(\mathrm{x}\right)+2\mathrm{y}\left(\mathrm{x}\right)=0$

 

The auxiliary equation for the above equation,

 

${\mathrm{m}}^2-3\mathrm{m}+2=0$

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

Solve the auxiliary equation,

 $\begin{array}{c}{\mathrm{m}}^2-3\mathrm{m}+2=0\\ {\mathrm{m}}^2-2\mathrm{m}-\mathrm{m}+2=0\\ \mathrm{m}(\mathrm{m}-2)-1(\mathrm{m}-2)=0\\ (\mathrm{m}-1)(\mathrm{m}-2)=0\end{array}$

 

The roots of the auxiliary equation are, 

 ${\mathrm{m}}_1=1,\&{\mathrm{m}}_2=2$

 

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{2\mathrm{x}}$

Step 3: Use the method of undetermined coefficients

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

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

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 $\mathbf{\alpha }\mathbf{+}\mathbf{i\beta }$ is a root of the associated auxiliary equation.

 

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

 

Assume, the particular solution of equation (1),

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^{\mathrm{s}}({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_1\mathrm{t}+{\mathrm{A}}_0){\mathrm{e}}^{\mathrm{\alpha x}}\mathrm{cos\beta x}+{\mathrm{t}}^{\mathrm{s}}({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_1\mathrm{t}+{\mathrm{B}}_0){\mathrm{e}}^{\mathrm{\alpha x}}\mathrm{sin\beta x}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{t}}^0\left({\mathrm{A}}_0\right){\mathrm{e}}^{\left(1\right)\mathrm{x}}\cos \left(1\right)\mathrm{x}+{\mathrm{t}}^0\left({\mathrm{B}}_0\right){\mathrm{e}}^{\left(1\right)\mathrm{x}}\sin \left(1\right)\mathrm{x}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{A}}_0{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}+{\mathrm{B}}_0{\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{x}}(\mathrm{Acosx}+\mathrm{Bsinx})\dots \left(3\right)\end{array}$ 

 

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

 $\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{x}}(\mathrm{Acosx}+\mathrm{Bsinx})+{\mathrm{e}}^{\mathrm{x}}(-\mathrm{Asinx}+\mathrm{Bcosx})\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)=(\mathrm{B}-\mathrm{A}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+(\mathrm{A}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=(\mathrm{B}-\mathrm{A}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+(\mathrm{B}-\mathrm{A}){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}+(\mathrm{A}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}-(\mathrm{A}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\\ {\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)=(-2\mathrm{A}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+\left(2\mathrm{B}\right){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}\end{array}$

 

Substitute the value of  ${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right),{\mathrm{y}}_{\mathrm{p}}\text{'}\left(\mathrm{t}\right)$ and ${\mathrm{y}}_{\mathrm{p}}\text{'}\text{'}\left(\mathrm{t}\right)$ the equation (1),

$\begin{array}{r}\Rightarrow \mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-3\mathrm{y}\text{'}\left(\mathrm{x}\right)+2\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\\ \Rightarrow -\left(2\mathrm{A}\right){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+\left(2\mathrm{B}\right){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}-3[(\mathrm{B}-\mathrm{A}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+(\mathrm{A}+\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}]+2\left[{\mathrm{e}}^{\mathrm{x}}(\mathrm{Acosx}+\mathrm{Bsinx})\right]\\ ={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\\ \Rightarrow (\mathrm{A}-\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}+(-\mathrm{A}-\mathrm{B}){\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}={\mathrm{e}}^{\mathrm{x}}\mathrm{sinx}\end{array}$

Comparing all coefficients of the above equation,

 

$\begin{array}{c}\mathrm{A}-\mathrm{B}=1......\left(4\right)\\ -\mathrm{A}-\mathrm{B}=0\end{array}$

 

Solve the above equations,

$\begin{array}{l} \mathrm{A}-\mathrm{B}=1\\ - \mathrm{A}-\mathrm{B}=0\\ \overline{\begin{array}{l}-2\mathrm{B}=1\\ \text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{B}=\frac{-1}{2}\end{array}}\end{array}$

Substitute the value of B in the equation (4),

  $\begin{array}{c}\mathrm{A}-\left(\frac{-1}{2}\right)=1\\ \mathrm{A}=\frac{1}{2}\end{array}$

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

$\begin{array}{c}{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{x}}(\mathrm{Acosx}+\mathrm{Bsinx})\\ {\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{x}}\left(\frac{1}{2}\mathrm{cosx}-\frac{1}{2}\mathrm{sinx}\right)\end{array}$

Therefore, the particular solution of equation (1),

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)={\mathrm{e}}^{\mathrm{x}}\left(\frac{1}{2}\mathrm{cosx}-\frac{1}{2}\mathrm{sinx}\right)$

Step 5: Conclusion

Therefore, the general solution is,

$\begin{array}{c}\mathrm{y}={\mathrm{y}}_{\mathrm{c}}\left(\mathrm{t}\right)+{\mathrm{y}}_{\mathrm{p}}\left(\mathrm{t}\right)\\ \mathrm{y}={\mathrm{c}}_1{\mathrm{e}}^{\mathrm{x}}+{\mathrm{c}}_2{\mathrm{e}}^{2\mathrm{x}}+\frac{1}{2}{\mathrm{e}}^{\mathrm{x}}(\mathrm{cosx}-\mathrm{sinx})\end{array}$