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Expert-verified Found in: Page 180 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Decide whether or not the method of undetermined coefficients can be applied to find a particular solution of the given equation.${\mathbf{2}}{\mathbf{y}}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\left(\mathbf{x}\right){\mathbf{-}}{\mathbf{6}}{\mathbf{y}}{\mathbf{\text{'}}}\left(\mathbf{x}\right){\mathbf{+}}{\mathbf{y}}\left(\mathbf{x}\right){\mathbf{=}}\frac{\mathrm{sinx}}{{\mathbf{e}}^{4x}}$

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

See the step by step solution

## Step 1: Use the method of undetermined coefficients

Given equation,

$2\mathrm{y}\text{'}\text{'}\left(\mathrm{x}\right)-6\mathrm{y}\text{'}\left(\mathrm{x}\right)+\mathrm{y}\left(\mathrm{x}\right)=\frac{\mathrm{sin}x}{{\mathrm{e}}^{4\mathrm{x}}}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}.....\left(1\right)$

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,

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

The roots of the auxiliary equation are,

${\mathrm{m}}_{1}=\frac{3+\sqrt{7}}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=\frac{3-\sqrt{7}}{2}$

The complementary solution of the given equation is,

${\mathrm{y}}_{\mathrm{c}}\left(\mathrm{x}\right)={\mathrm{e}}^{\frac{3}{2}\mathrm{x}}\left[{\mathrm{c}}_{1}\mathrm{cos}\left(\frac{\sqrt{7}}{2}\right)+{\mathrm{c}}_{2}\mathrm{sin}\left(\frac{\sqrt{7}}{2}\right)\right]$

## 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,

$\mathbf{\alpha }\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{,}\text{\hspace{0.17em}\hspace{0.17em}}\mathbf{\beta }\mathbf{=}\mathbf{1}$

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,

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)={\mathrm{t}}^{\mathrm{s}}\left({\mathrm{A}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{A}}_{1}\mathrm{t}+{\mathrm{A}}_{0}\right){\mathrm{e}}^{\mathrm{\alpha x}}\mathrm{cos\beta x}+{\mathrm{t}}^{\mathrm{s}}\left({\mathrm{B}}_{\mathrm{m}}{\mathrm{t}}^{\mathrm{m}}+...+{\mathrm{B}}_{1}\mathrm{t}+{\mathrm{B}}_{0}\right){\mathrm{e}}^{\mathrm{\alpha x}}\mathrm{sin\beta x}$

Therefore, the method of undetermined coefficients can be applied. ### Want to see more solutions like these? 