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Found in: Page 249

### 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

# Let ${\mathbf{A}}{\mathbf{=}}{\mathbf{D}}{\mathbf{-}}{\mathbf{1}}{\mathbf{,}}{\mathbf{B}}{\mathbf{=}}{\mathbf{D}}{\mathbf{+}}{\mathbf{2}}{\mathbf{,}}{\mathbf{C}}{\mathbf{=}}{{\mathbf{D}}}^{{\mathbf{2}}}{\mathbf{+}}{\mathbf{D}}{\mathbf{-}}{\mathbf{2}}{\mathbf{,}}$ where ${\mathbf{D}}{\mathbf{=}}\frac{\mathbf{d}}{\mathbf{dt}}$. For ${\mathbf{y}}{\mathbf{=}}{{\mathbf{t}}}^{{\mathbf{3}}}{\mathbf{-}}{\mathbf{8}}$, compute (a) ${\mathbf{A}}\left[\mathbf{y}\right]$(b) ${\mathbf{B}}\left[\mathbf{A}\left[\mathbf{y}\right]\right]$(c) ${\mathbf{B}}\left[\mathbf{y}\right]$(d) ${\mathbf{A}}\left[\mathbf{B}\left[\mathbf{y}\right]\right]$(e) ${\mathbf{C}}\left[\mathbf{y}\right]$

(a) The solution of $\mathbf{A}\left[\mathbf{y}\right]$ is $\mathbf{-}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{8}$.

(b) The solution of $\mathbf{B}\left[\mathbf{A}\left[\mathbf{y}\right]\right]$ is $\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{t}\mathbf{+}\mathbf{16}$

(c) The solution of $\mathbf{B}\left[\mathbf{y}\right]$ is $\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{16}$

(d) The solution of $\mathbf{A}\left[\mathbf{B}\left[\mathbf{y}\right]\right]$ is $\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{t}\mathbf{+}\mathbf{16}$

(e) The solution of $\mathbf{C}\left[\mathbf{y}\right]$ is $\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{t}\mathbf{+}\mathbf{16}$

See the step by step solution

## Step 1: General form

Elimination Procedure for 2 × 2 Systems:

To find a general solution for the system

${\mathbf{L}}_{\mathbf{1}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{2}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{1}}\mathbf{,}\phantom{\rule{0ex}{0ex}}{\mathbf{L}}_{\mathbf{3}}\left[\mathbf{x}\right]\mathbf{+}{\mathbf{L}}_{\mathbf{4}}\left[\mathbf{y}\right]\mathbf{=}{\mathbf{f}}_{\mathbf{2}}\mathbf{,}\phantom{\rule{0ex}{0ex}}$

a. Make sure that the system is written in operator form.

b. Eliminate one of the variables, say, y, and solve the resulting equation for x(t).

If the system is degenerate, stop! A separate analysis is required to determine whether or not there are solutions.

c. (Shortcut) If possible, use the system to derive an equation that involves y(t) but not its derivatives. [Otherwise, go to step (d).] Substitute the found expression for x(t) into this equation to get a formula for y(t). The expressions for x(t), y(t) gives the desired general solution.

d. Eliminate x from the system and solve for y(t). [Solving for y(t) gives more constants----in fact, twice as many as needed.]

e. Remove the extra constants by substituting the expressions for x(t) and y(t) into one or both of the equations in the system. Write the expressions for x(t) and y(t) in terms of the remaining constants.

## Step 2: Evaluate the given equation

Given that, $\mathbf{A}\mathbf{=}\mathbf{D}\mathbf{-}\mathbf{1}\mathbf{,}\mathbf{B}\mathbf{=}\mathbf{D}\mathbf{+}\mathbf{2}\mathbf{,}\mathbf{C}\mathbf{=}{\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}\mathbf{,}$ where $\mathbf{D}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}$.

And $\mathbf{y}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}$.

Then compute the given parts,

$\mathbf{A}\left[\mathbf{y}\right]\mathbf{=}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}\left[{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}\right]\mathbf{-}\left[{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{8}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{-}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{8}$

## Step 3: Compute the parts (b) and (c)

$\mathbf{B}\left[\mathbf{A}\left[\mathbf{y}\right]\right]\mathbf{=}\left(\mathbf{D}\mathbf{+}\mathbf{2}\right)\left[\mathbf{-}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{8}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}\left[\mathbf{-}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{8}\right]\mathbf{+}\mathbf{2}\left[\mathbf{-}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{8}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{t}\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{6}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{16}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{t}\mathbf{+}\mathbf{16}$

$\mathbf{B}\left[\mathbf{y}\right]\mathbf{=}\left(\mathbf{D}\mathbf{+}\mathbf{2}\right)\left[{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}\left[{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}\right]\mathbf{+}\mathbf{2}\left[{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{16}$

## Step 4: Compute the parts (d) and (e)

$\mathbf{A}\left[\mathbf{B}\left[\mathbf{y}\right]\right]\mathbf{=}\left(\mathbf{D}\mathbf{-}\mathbf{1}\right)\left[\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{16}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\frac{\mathbf{d}}{\mathbf{dt}}\left[\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{16}\right]\mathbf{-}\left[\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{16}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{6}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{t}\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{16}\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{6}\mathbf{t}\mathbf{+}\mathbf{16}$

$\mathbf{C}\left[\mathbf{y}\right]\mathbf{=}\left({\mathbf{D}}^{\mathbf{2}}\mathbf{+}\mathbf{D}\mathbf{-}\mathbf{2}\right)\left[{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\frac{{\mathbf{d}}^{\mathbf{2}}}{{\mathbf{dt}}^{\mathbf{2}}}\left[{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}\right]\mathbf{+}\frac{\mathbf{d}}{\mathbf{dt}}\left[{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}\right]\mathbf{-}\mathbf{2}\left[{\mathbf{t}}^{\mathbf{3}}\mathbf{-}\mathbf{8}\right]\phantom{\rule{0ex}{0ex}}\mathbf{=}\mathbf{6}\mathbf{t}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{-}\mathbf{2}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{16}$

So, the solutions are founded.

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