Q3E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation.$\mathbf{2}\mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{-}\mathbf{6}\mathbf{y}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{+}\mathbf{y}\left(\mathbf{x}\right)\mathbf{=}\frac{\mathrm{sinx}}{{\mathbf{e}}^{4x}}$

Step-by-Step Solution

Answer

Yes, the method of undetermined coefficients can be applied to find a particular solution of the given equation.

Step 1: Use the method of undetermined coefficients

Given equation,

$2\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-6\mathrm{y}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)=\frac{\sin x}{{\mathrm{e}}^{4\mathrm{x}}}.....\left(1\right)$

 

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

 

$2\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-6\mathrm{y}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)=0$

 

The auxiliary equation for the above equation,

 

$2{\mathrm{m}}^2-6\mathrm{m}+1=0$

 

Step 2: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

 

$\begin{array}{c}2{\mathrm{m}}^2-6\mathrm{m}+1=0\\ \mathrm{m}=\frac{-(-6)\pm \sqrt{36-4\left(2\right)\left(1\right)}}{2\left(2\right)}\\ \mathrm{m}=\frac{6\pm \sqrt{28}}{4}\\ \mathrm{m}=\frac{3\pm \sqrt{7}}{2}\end{array}$

 

The roots of the auxiliary equation are, 

 ${\mathrm{m}}_1=\frac{3+\sqrt{7}}{2},{\mathrm{m}}_2=\frac{3-\sqrt{7}}{2}$

 

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)={\mathrm{e}}^{\frac{3}{2}\mathrm{x}}\left[{\mathrm{c}}_1\cos \left(\frac{\sqrt{7}}{2}\right)+{\mathrm{c}}_2\sin \left(\frac{\sqrt{7}}{2}\right)\right]$

Step 3: Final conclusion.

According to the method of undetermined coefficients, 

 

If $\mathrm{\beta }\ne 0$, then the blow equation has a particular solution,

 

$\mathrm{ay}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{by}\text{'}\left(\mathrm{x}\right)+\mathrm{cy}\left(\mathrm{x}\right)={\mathrm{Ct}}^{\mathrm{m}}{\mathrm{e}}^{\mathrm{\alpha x}}\mathrm{sin\beta x}$

 

Compare with the given differential equation,

 $2\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-6\mathrm{y}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)={\mathrm{e}}^{-4\mathrm{x}}\mathrm{sinx}$

 We have,

$\mathbf{\alpha }\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{,}\mathbf{\beta }\mathbf{=}\mathbf{1}$

Condition satisfies,

 

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

 

The roots of the auxiliary equation are different,

 

Therefore, s=0  

 

The particular solution of the equation,

 

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

 

Therefore, the method of undetermined coefficients can be applied.