 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q6E

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.${\mathbf{2}}{\mathbf{\omega }}{\mathbf{\text{'}}}{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{-}}{\mathbf{3}}{\mathbf{\omega }}\mathbf{\left(}\mathbf{x}\mathbf{\right)}{\mathbf{=}}{\mathbf{4}}{{\mathbf{xsin}}}^{2}{\mathbf{x}}{\mathbf{+}}{\mathbf{4}}{{\mathbf{xcos}}}^{2}{\mathbf{x}}$

Yes, the method of undetermined coefficients can be applied.

See the step by step solution

## Step 1: Simplification of the given differential equation.

Given equation,

$2\mathrm{\omega }\text{'}\text{'}\left(\mathrm{x}\right)-3\mathrm{\omega }\left(\mathrm{x}\right)=4{\mathrm{xsin}}^{2}\mathrm{x}+4{\mathrm{xcos}}^{2}\mathrm{x}$

Simplify the above equation,

$\begin{array}{l}2\mathrm{\omega }\text{'}\text{'}\left(\mathrm{x}\right)-3\mathrm{\omega }\left(\mathrm{x}\right)=4\mathrm{x}\left({\mathrm{sin}}^{2}\mathrm{x}+{\mathrm{cos}}^{2}\mathrm{x}\right)\\ 2\mathrm{\omega }\text{'}\text{'}\left(\mathrm{x}\right)-3\mathrm{\omega }\left(\mathrm{x}\right)=4\mathrm{x}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}......\left(1\right)\end{array}$

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

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

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

The auxiliary equation for the above equation,

role="math" localid="1654859563585" $\begin{array}{c}2{\mathrm{m}}^{2}-3=0\\ 2{\mathrm{m}}^{2}=3\\ \mathrm{m}=±\frac{3}{2}\end{array}$

The roots of the auxiliary equation are,

role="math" localid="1654859598798" ${\mathrm{m}}_{1}=\frac{3}{2},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}{\mathrm{m}}_{2}=-\frac{3}{2}$

The complementary solution of the given equation is,

${\mathbf{\omega }}_{c}\left(\mathbf{x}\right)\mathbf{=}{\mathbf{c}}_{1}{\mathbf{e}}^{\frac{3}{2}\mathbf{x}}\mathbf{+}{\mathbf{c}}_{2}{\mathbf{e}}^{\mathbf{-}\frac{3}{2}\mathbf{x}}$.

## Step 3: Final conclusion

The R.H.S. of equation is (4x).

Therefore, the particular solution of the equation,

${\mathrm{y}}_{\mathrm{p}}\left(\mathrm{x}\right)=\mathrm{Ax}+\mathrm{b}$

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