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Q38E

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Found in: Page 338

### 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 38 and 39, use the elimination method of Section to find a general solution to the given system.${\mathbit{x}}{\mathbf{-}}{{\mathbit{d}}}^{{\mathbf{2}}}{\mathbit{y}}{\mathbf{/}}{\mathbit{d}}{{\mathbit{t}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbit{t}}{\mathbf{+}}{\mathbf{1}}$${\mathbit{d}}{\mathbit{x}}{\mathbf{/}}{\mathbit{d}}{\mathbit{t}}{\mathbf{+}}{\mathbit{d}}{\mathbit{y}}{\mathbf{/}}{\mathbit{d}}{\mathbit{t}}{\mathbf{-}}{\mathbf{2}}{\mathbit{y}}{\mathbf{=}}{{\mathbit{e}}}^{{\mathbf{t}}}$

The general solution is $x\left(t\right)=\frac{-3{c}_{1}-\sqrt{7}{c}_{2}}{2}{e}^{-\frac{t}{2}}\mathrm{cos}\frac{\sqrt{7}}{2}t+\frac{\sqrt{7}{c}_{1}-3{c}_{2}}{2}{e}^{-\frac{t}{2}}\mathrm{sin}\frac{\sqrt{7}}{2}t+\left({c}_{3}+\frac{1}{2}\right){e}^{t}+\frac{1}{4}t{e}^{t}+t+1\phantom{\rule{0ex}{0ex}}y\left(t\right)={c}_{1}{e}^{-\frac{t}{2}}\mathrm{cos}\frac{\sqrt{7}}{2}t+{c}_{2}{e}^{-\frac{t}{2}}\mathrm{sin}\frac{\sqrt{7}}{2}t+{c}_{3}{e}^{t}+\frac{1}{4}t{e}^{t}+1$

See the step by step solution

## Step 1: Definition

A differential equation is an equation that contains one or more functions with its derivatives.

## Step 2: Simplify equation

It is given that $x-{y}^{\text{'}\text{'}}=t+1$

$⇒x={y}^{\text{'}\text{'}}+t+1$-------(1)

Differentiating equation (1) we have:

${x}^{\text{'}}={y}^{\text{'}\text{'}\text{'}}+1$

Substituting value we get:

${x}^{\text{'}}+{y}^{\text{'}}-2y={e}^{t}⇒{y}^{\text{'}\text{'}\text{'}}+{y}^{\text{'}}-2y+1={e}^{\text{'}}$

## Step 3: For general solution

The auxiliary equation is given by:

$\left({D}^{3}+D-2\right)y={e}^{t}-1$

The homogenous equation is ${D}^{3}+D-2=0$:

$D=1,\frac{-1}{2}±i\frac{\sqrt{7}}{2}$

So, the general solution is given by ${y}_{g}={C}_{1}{e}^{t}+{C}_{2}{e}^{\frac{-t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}}{2}t\right)+{C}_{3}{e}^{\frac{-t}{2}}\mathrm{sin}\left(\frac{\sqrt{7}t}{2}\right)$

## Step 4: For particular solution

Let ${y}_{p}=At{e}^{t}+B$

Differentiating we have:

${y}_{p}^{\text{'}}=A{e}^{t}\left(t+1\right)\phantom{\rule{0ex}{0ex}}{y}_{p}^{\text{'}\text{'}}=A{e}^{t}\left(t+2\right)$

${y}_{p}^{\text{'}\text{'}\text{'}}=A{e}^{t}\left(t+3\right)\phantom{\rule{0ex}{0ex}}{y}_{p}^{\text{'}\text{'}\text{'}}+{y}_{p}^{\text{'}}-2{y}_{p}={e}^{t}-1$

Substituting value & comparing we get:

$A{e}^{t}\left(t+3+t+1-2t\right)-2B={e}^{t}-1\phantom{\rule{0ex}{0ex}}4A=1,-2B=-1\phantom{\rule{0ex}{0ex}}A=\frac{1}{4},B=\frac{1}{2}$

So $y={C}_{1}{e}^{t}+{C}_{2}{e}^{\frac{-t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}}{2}t\right)+{C}_{3}{e}^{\frac{-t}{2}}\mathrm{sin}\left(\frac{\sqrt{7}}{2}t\right)+\frac{t{e}^{t}}{4}+\frac{1}{2}$

## Step 5: Compute value x

Now we need to substitute value of $y\text{'}\text{'}$in $x={y}^{\text{'}\text{'}}+t+1$.

${y}^{\text{'}}={C}_{1}{e}^{\text{'}}-\frac{{C}_{2}}{2}{e}^{\frac{-t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}t}{2}\right)-\frac{\sqrt{7}}{2}{C}_{2}{e}^{\frac{-t}{2}}\mathrm{sin}\left(\frac{\sqrt{7}t}{2}\right)+\frac{\sqrt{7}}{2}{C}_{3}{e}^{\frac{-t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}t}{2}\right)-\frac{{C}_{3}}{2}{e}^{\frac{-t}{2}}\mathrm{sin}\left(\frac{\sqrt{7}t}{2}\right)+\frac{{e}^{t}}{4}\left(t+1\right)\phantom{\rule{0ex}{0ex}}{y}^{\text{'}}={e}^{\frac{-t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}t}{2}\right)\left(\frac{\sqrt{7}{C}_{3}-{C}_{2}}{2}\right)+{e}^{\frac{-t}{2}}\mathrm{sin}\left(\frac{\sqrt{7}t}{2}\right)\left(\frac{-\sqrt{7}{C}_{2}-{C}_{3}}{2}\right)+{e}^{t}\left(\frac{t}{4}+{C}_{1}+\frac{1}{4}\right)\phantom{\rule{0ex}{0ex}}$

And

${y}^{\text{'}\text{'}}={e}^{\frac{-t}{2}}\mathrm{sin}\left(\frac{\sqrt{7}t}{2}\right)\left(\frac{-{C}_{2}+{C}_{3}\sqrt{7}}{2}\right)\left(\frac{-\sqrt{7}}{2}\right)-\frac{1}{2}{e}^{\frac{t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}t}{2}\right)\left(\frac{\sqrt{7}{C}_{3}-{C}_{2}}{2}\right)-\left(\frac{\sqrt{7}{C}_{2}+{C}_{3}}{2}\right)\left\{\frac{\sqrt{7}}{2}{e}^{\frac{-t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}t}{2}\right)-\frac{1}{2}{e}^{\frac{-t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}t}{2}\right)\right\}+{e}^{t}\left(\frac{t}{4}+{C}_{1}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}{y}^{\text{'}\text{'}}={e}^{\frac{-t}{2}}\mathrm{sin}\left(\frac{\sqrt{7}t}{2}\right)\left(\frac{\sqrt{7}{C}_{2}-3{C}_{3}}{4}\right)+{e}^{\frac{-t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}t}{2}\right)\left(\frac{-3{C}_{2}-\sqrt{7}{C}_{3}}{2}\right)+{e}^{t}\left(\frac{t}{4}+{C}_{1}+\frac{1}{2}\right)\phantom{\rule{0ex}{0ex}}$

Substituting values we get:

$x={e}^{\frac{-t}{2}}\mathrm{sin}\left(\frac{\sqrt{7}t}{2}\right)\left(\frac{\sqrt{7}{C}_{2}-3{C}_{3}}{4}\right)+{e}^{\frac{-t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}t}{2}\right)\left(\frac{-3{C}_{2}-\sqrt{7}{C}_{3}}{2}\right)+{e}^{t}\left(\frac{t}{4}+{C}_{1}+\frac{1}{2}\right)+t+1$

Hence the solution of equation is given by

$x={e}^{\frac{-t}{2}}\mathrm{sin}\left(\frac{\sqrt{7}t}{2}\right)\left(\frac{\sqrt{7}{C}_{2}-3{C}_{3}}{4}\right)+{e}^{\frac{-t}{2}}\mathrm{cos}\left(\frac{\sqrt{7}t}{2}\right)\left(\frac{-3{C}_{2}-\sqrt{7}{C}_{3}}{2}\right)+{e}^{t}\left(\frac{t}{4}+{C}_{1}+\frac{1}{2}\right)+t+1$

Therefore the solution is :

$x\left(t\right)=\frac{-3{c}_{1}-\sqrt{7}{c}_{2}}{2}{e}^{-\frac{t}{2}}\mathrm{cos}\frac{\sqrt{7}}{2}t+\frac{\sqrt{7}{c}_{1}-3{c}_{2}}{2}{e}^{-\frac{t}{2}}\mathrm{sin}\frac{\sqrt{7}}{2}t+\left({c}_{3}+\frac{1}{2}\right){e}^{t}+\frac{1}{4}t{e}^{t}+t+1\phantom{\rule{0ex}{0ex}}y\left(t\right)={c}_{1}{e}^{-\frac{t}{2}}\mathrm{cos}\frac{\sqrt{7}}{2}t+{c}_{2}{e}^{-\frac{t}{2}}\mathrm{sin}\frac{\sqrt{7}}{2}t+{c}_{3}{e}^{t}+\frac{1}{4}t{e}^{t}+1$