 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q10E

Expert-verified Found in: Page 191 ### 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 # In Problems 9 and 10, find a particular solution first by undetermined coefficients, and then by variation of parameters. Which method was quicker?${\mathbf{10}}\mathbf{.}\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{2}{{\mathbit{e}}}^{\mathbf{2}\mathbf{t}}$

The general solution is $y\left(t\right)={c}_{1}{e}^{2t}+{c}_{2}2{e}^{-t}+\frac{t}{3}{e}^{2t}-\frac{1}{9}{e}^{2t}$.

See the step by step solution

## Step 1: Find a particular solution by variation of parameter.

The differential equation is $2x\text{'}\text{'}\left(t\right)-2x\text{'}\left(t\right)-4x\left(t\right)=2{e}^{2t}$

This can be written as $x\text{'}\text{'}\left(t\right)-x\text{'}\left(t\right)-2x\left(t\right)={e}^{2t}$

The homogenous equation is ${r}^{2}-r-2=0$.

Two independent solutions are $r=2,-1$.

Then ${y}_{1}={e}^{2t},{y}_{2}={e}^{-t}$

${y}_{h}\left(t\right)={c}_{1}{e}^{2t}+{c}_{2}{e}^{-t}$

The particular solution is ${y}_{p}={v}_{1}\left(t\right){e}^{2t}+{v}_{2}\left(t\right){e}^{-t}$.

## Step 2: Evaluate, v1 and v2, v'1 and v1, v'2 and v2

Here ${y}_{p}={v}_{1}\left(t\right){e}^{t}+{v}_{2}\left(t\right)t{e}^{-t}$

And referring to (9) $y\left(t\right)={c}_{1}{e}^{\alpha t}\mathrm{cos}\beta t{c}_{2}{e}^{\alpha t}\mathrm{sin}\beta t$ and solve the system by derivative then,

${v}_{1}\text{'}{e}^{2t}+{v}_{2}\text{'}{e}^{-t}=0\phantom{\rule{0ex}{0ex}}2{v}_{1}\text{'}{e}^{2t}-{v}_{2}\text{'}{e}^{-t}=\frac{f}{a}\phantom{\rule{0ex}{0ex}}2{v}_{1}\text{'}{e}^{2t}-{v}_{2}\text{'}{e}^{-t}={e}^{2t}$

Now for finding the values.

${v}_{1}\text{'}=\frac{-f\left(t\right){y}_{2}\left(t\right)}{a\left[{y}_{1}\left(t\right)y{\text{'}}_{2}\left(t\right)-y{\text{'}}_{1}\left(t\right){y}_{2}\left(t\right)\right]}\phantom{\rule{0ex}{0ex}}=\frac{-{e}^{2t}.{e}^{-t}}{\left[-{e}^{2t}.{e}^{-t}-2{e}^{-t}.{e}^{-t}\right]}\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$

Now integrating this;

${v}_{1}\left(t\right)=\int \frac{1}{3} dt\phantom{\rule{0ex}{0ex}}=\frac{t}{3}+C\phantom{\rule{0ex}{0ex}}{v}_{2}\text{'}=\frac{f\left(t\right){y}_{1}\left(t\right)}{a\left[{y}_{1}\left(t\right)y{\text{'}}_{2}\left(t\right)-y{\text{'}}_{1}\left(t\right){y}_{2}\left(t\right)\right]}\phantom{\rule{0ex}{0ex}}=\frac{{e}^{2t}.{e}^{t}}{\left[-{e}^{2t}.{e}^{-t}-2{e}^{-t}.{e}^{-t}\right]}\phantom{\rule{0ex}{0ex}}=-\frac{1}{3}{e}^{3t}$

Integrate this.

${v}_{2}\left(t\right)=\int -\frac{1}{3}{e}^{3t}dt\phantom{\rule{0ex}{0ex}}=-\frac{1}{9}{e}^{3t}+C$

Thus, the particular solution is when $C=0$

${y}_{p}=\frac{t}{3}{e}^{2t}+C-\left(\frac{1}{9}{e}^{3t}+C\right){e}^{-t}\phantom{\rule{0ex}{0ex}}{y}_{p}=\frac{t}{3}{e}^{2t}-\frac{1}{9}{e}^{2t}$

Therefore, the general solution is:

$y\left(t\right)={y}_{h}\left(t\right)+{y}_{p}\left(t\right)\phantom{\rule{0ex}{0ex}}y\left(t\right)={c}_{1}{e}^{2t}+{c}_{2}2{e}^{-t}+\frac{t}{3}{e}^{2t}-\frac{1}{9}{e}^{2t}$ ### Want to see more solutions like these? 