# 2nd Order diffn eqn using Euler's eqn

• Nov 18th 2006, 12:13 PM
ashura
2nd Order diffn eqn using Euler's eqn
Need help to find the particular integral $y_p$ of a 2nd order diffn eqn using Euler's eqn $only$;
I know that there are other methods of solution. Just want to understand the principle (Euler's) that i'm trying to apply to my type of solution.

$\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y = 2sin4x$

$D=\frac{d}{dx}, y_p =$ particular integral,

$(D^2-5D+6)y_p = 2sin4x$

$y_p \times [(D-2)(D-3)] = 2sin4x$ ,

$\therefore y_p = \frac{2sin4x}{(D-2)(D-3)}$

By Euler's eqn $e^{4ix} = cos4x + isin4x$

This is where i'm uncertain,

$\therefore y_p = Real[ \frac{e^{4ix}}{(D-2)(D-3)}] + 2\times Imaginary[\frac{e^{4ix}}{(D-2)(D-3)}]$

$2\times Imaginary[\frac{e^{4ix}}{(D-2)(D-3)}]$ because question has $2sin4x$ on RHS

Question:
1) Should the Real part be included seeing that the question has the imaginary part (sin4x) only?

next, Remember $D = \frac{d}{dx}$

$\therefore y_p = Real \frac{e^{4ix}}{(4i-2)(4i-3)}] + 2\times Imaginary[\frac{e^{4ix}}{(4i-2)(4i-3)}]$ - Replace D with coefficient of x in $e^{4ix}$
My final answer wasn't correct. Seek help using my method and not another approach.
Thanks.
• Nov 18th 2006, 12:54 PM
topsquark
Quote:

Originally Posted by ashura
Need help to find the particular integral $y_p$ of a 2nd order diffn eqn using Euler's eqn $only$;
I know that there are other methods of solution. Just want to understand the principle (Euler's) that i'm trying to apply to my type of solution.

$\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y = 2sin4x$

$D=\frac{d}{dx}, y_p =$ particular integral,

$(D^2-5D+6)y_p = 2sin4x$

$y_p \times [(D-2)(D-3)] = 2sin4x$ ,

$\therefore y_p = \frac{2sin4x}{(D-2)(D-3)}$

By Euler's eqn $e^{4ix} = cos4x + isin4x$

This is where i'm uncertain,

$\therefore y_p = Real[ \frac{e^{4ix}}{(D-2)(D-3)}] + 2\times Imaginary[\frac{e^{4ix}}{(D-2)(D-3)}]$

$2\times Imaginary[\frac{e^{4ix}}{(D-2)(D-3)}]$ because question has $2sin4x$ on RHS

Question:
1) Should the Real part be included seeing that the question has the imaginary part (sin4x) only?

next, Remember $D = \frac{d}{dx}$

$\therefore y_p = Real \frac{e^{4ix}}{(4i-2)(4i-3)}] + 2\times Imaginary[\frac{e^{4ix}}{(4i-2)(4i-3)}]$ - Replace D with coefficient of x in $e^{4ix}$
My final answer wasn't correct. Seek help using my method and not another approach.
Thanks.

I should mention I'm VERY rusty with this as it's been a long time since I've seen it.

1)
$(D^2-5D+6)y_p = 2sin4x$

$y_p \times [(D-2)(D-3)] = 2sin4x$

Don't write your second step. The operator (D-2)(D-3) needs something to act on, so the $y_p$ can't come out on the left. You aren't really multiplying here and when you move the operator to the RHS you aren't really dividing. You are operating on both sides of the equation with $[(D-2)(D-3)]^{-1}$ which we write in shorthand as $\frac{1}{(D-2)(D-3)}$.

2) Yes, drop the real part of the expression.

-Dan
• Nov 18th 2006, 03:18 PM
ashura
$\therefore y_p = 2\times Imaginary[\frac{e^{4ix}}{(4i-2)(4i-3)}]$ - Replace D with coefficient of x in $e^{4ix}$

$\therefore y_p = 2\times Imaginary[\frac{e^{4ix}}{(16i^2-12i-8i+6)}]$

$\therefore y_p = 2\times Imaginary[\frac{e^{4ix}}{(6-36i)}]$

$\therefore y_p = \frac{2}{6}\times Imaginary[\frac{e^{4ix}}{(-36i)}]$

$\therefore y_p = \frac{1}{3}(\frac{sin4x}{-36i})$

That's not the right answer which has cos4x in it. Where did I go wrong?
• Nov 18th 2006, 05:31 PM
topsquark
Quote:

Originally Posted by ashura
$\therefore y_p = 2\times Imaginary[\frac{e^{4ix}}{(4i-2)(4i-3)}]$ - Replace D with coefficient of x in $e^{4ix}$

$\therefore y_p = 2\times Imaginary[\frac{e^{4ix}}{(16i^2-12i-8i+6)}]" alt="\therefore y_p = 2\times Imaginary[\frac{e^{4ix}}{(16i^2-12i-8i+6)}]" />

$\therefore y_p = 2\times Imaginary[\frac{e^{4ix}}{(6-36i)}]$

$\therefore y_p = \frac{2}{6}\times Imaginary[\frac{e^{4ix}}{(-36i)}]$

$\therefore y_p = \frac{1}{3}(\frac{sin4x}{-36i})$

That's not the right answer which has cos4x in it. Where did I go wrong?

$y_p = 2 \mathbb{IM} \left ( \frac{e^{4ix}}{(4i - 2)(4i - 3)} \right ) = \frac{2}{25}cos(4x) - \frac{1}{25}sin(4x)$

In line 3, that should be "-10 - 20i," not "6 - 36i."

In line 4 you need to rationalize the denominator, not just drop the real part. Once you have rationalized, then use [tex]e^{4ix} = cos(4x) + i \, sin(4x), then expand the product, THEN take the imaginary part.

$y_p = 2 \mathbb{IM} \left ( \frac{e^{4ix}}{(4i - 2)(4i - 3)} \right ) = 2 \mathbb{IM} \left (e^{4ix} \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right ) \right )$

= $2 \mathbb{IM} \left ( cos(4x) + i \, sin(4x) \right ) \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right )$

= $2 \mathbb{IM} \left ( -cos(4x) \cdot \frac{1}{50} - i \, sin(4x) \cdot \frac{1}{50} + cos(4x) \cdot \frac{1}{25}i + i \, sin(4x) \cdot \frac{1}{25}i \right )$

= $2 \left ( \frac{1}{25}cos(4x) - \frac{1}{50}sin(4x) \right )$

= $\frac{2}{25}cos(4x) - \frac{1}{25}sin(4x)$

-Dan
• Nov 19th 2006, 06:30 AM
ashura
Quote:

Originally Posted by topsquark
$y_p = 2 \mathbb{IM} \left ( \frac{e^{4ix}}{(4i - 2)(4i - 3)} \right ) = \frac{2}{25}cos(4x) - \frac{1}{25}sin(4x)$

In line 3, that should be "-10 - 20i," not "6 - 36i."

In line 4 you need to rationalize the denominator, not just drop the real part. Once you have rationalized, then use [tex]e^{4ix} = cos(4x) + i \, sin(4x), then expand the product, THEN take the imaginary part.

$y_p = 2 \mathbb{IM} \left ( \frac{e^{4ix}}{(4i - 2)(4i - 3)} \right ) = 2 \mathbb{IM} \left (e^{4ix} \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right ) \right )$

= $2 \mathbb{IM} \left ( cos(4x) + i \, sin(4x) \right ) \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right )$

= $2 \mathbb{IM} \left ( -cos(4x) \cdot \frac{1}{50} - i \, sin(4x) \cdot \frac{1}{50} + cos(4x) \cdot \frac{1}{25}i + i \, sin(4x) \cdot \frac{1}{25}i \right )$

= $2 \left ( \frac{1}{25}cos(4x) - \frac{1}{50}sin(4x) \right )$

= $\frac{2}{25}cos(4x) - \frac{1}{25}sin(4x)$

-Dan

Thanks for the correction. You got the right answer and I follow your working. Now cos4x is usually the real part. Did you 2IM both cos4x and sin4x because the given equation had only the imaginary or sin part?: =

= $2 \mathbb{IM} \left ( cos(4x) + i \, sin(4x) \right ) \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right )$

So that if the question had cos4x + 2sin4x on the RHS you would have put: =

= $\mathbb{RE} \left ( cos(4x) \right ) + i \, 2 \mathbb{IM} \left ( sin(4x) \right ) \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right )$
• Nov 19th 2006, 11:05 AM
topsquark
Quote:

Originally Posted by ashura
Thanks for the correction. You got the right answer and I follow your working. Now cos4x is usually the real part. Did you 2IM both cos4x and sin4x because the given equation had only the imaginary or sin part?: =

= $2 \mathbb{IM} \left ( cos(4x) + i \, sin(4x) \right ) \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right )$

So that if the question had cos4x + 2sin4x on the RHS you would have put: =

= $\mathbb{RE} \left ( cos(4x) \right ) + i \, 2 \mathbb{IM} \left ( sin(4x) \right ) \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right )$

Note that I didn't take the imaginary part until I had multiplied the two terms out. This is because:
$\mathbb{IM}(a + ib)(c + id) \neq (b)(d) = bd$

$\mathbb{IM}(a + ib)(c + id) = \mathbb{IM}[(ac - bd) + i(ad + bc)] = ad + bc$

The product contains imaginary factors using both real components as well as the imaginary components, so we can't take the imaginary (or real) parts before we take the product. This is mainly what your conceptual error was in your previous derivation.

So what you would need to do the problem with the cos(4x) in it would be to take the real part of the whole expression, not just the cos(4x).

-Dan
• Nov 21st 2006, 06:06 PM
ashura
I think that I understand your point. I've worked the solution using cos4x + 2sin4x on the RHS. Thanks for your comment.

$\frac{d^2y}{dx^2}-5\frac{dy}{dx}+6y = cos4x + 2sin4x$

$y_p \{(D-2)(D-3) \} = cos4x + 2sin4x$

$\therefore y_p = \frac{cos4x + 2sin4x}{(D-2)(D-3)}$

By Euler's eqn $e^{4ix} = cos4x + isin4x$

$\therefore y_p = \mathbb {RE}[ \frac{e^{4ix}}{(D-2)(D-3)}] + 2 \mathbb {IM}[\frac{e^{4ix}}{(D-2)(D-3)}]$

$\therefore y_p = \mathbb {RE} [\frac{e^{4ix}}{(4i-2)(4i-3)}] + 2\mathbb {IM} [\frac{e^{4ix}}{(4i-2)(4i-3)}]$

$\therefore y_p = \mathbb {RE} [\frac{e^{4ix}}{(-10 - 20i}] + 2\mathbb {IM} [\frac{e^{4ix}}{(-10 -20i)}]$

$\therefore y_p = \mathbb {RE} [\frac{e^{4ix}.}{(-10 - 20i}] + 2\mathbb {IM} [\frac{e^{4ix}}{(-10 -20i)}]$

$y_p = \mathbb{RE} \left (e^{4ix} \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right ) \right ) +
2 \mathbb{IM}\left (e^{4ix} \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right ) \right )$

= $\mathbb{RE} \left ( cos(4x) + i \, sin(4x) \right ) \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right )$ + $2 \mathbb{IM} \left ( cos(4x) + i \, sin(4x) \right ) \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right )$

= $\mathbb{RE} \left ( -cos(4x) \cdot \frac{1}{50} - i \, sin(4x) \cdot \frac{1}{50} + cos(4x) \cdot \frac{1}{25}i + i \, sin(4x) \cdot \frac{1}{25}i \right )$ +
$2 \mathbb{IM} \left ( -cos(4x) \cdot \frac{1}{50} - i \, sin(4x) \cdot \frac{1}{50} + cos(4x) \cdot \frac{1}{25}i + i \, sin(4x) \cdot \frac{1}{25}i \right )$

= $\left ( -\frac{1}{50}cos(4x) - \frac{1}{25}sin(4x) \right ) + 2 \left ( \frac{1}{25}cos(4x) - \frac{1}{50}sin(4x) \right )$

= $-\frac{1}{25} \left ( \frac{cos(4x)}{2} + sin(4x) \right ) + \frac{2}{25}cos(4x) - \frac{1}{25}sin(4x)$