# Thread: 2nd Order diffn eqn using Euler's eqn

1. ## 2nd Order diffn eqn using Euler's eqn

Need help to find the particular integral $\displaystyle y_p$ of a 2nd order diffn eqn using Euler's eqn $\displaystyle only$;
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.

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

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

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

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

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

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

This is where i'm uncertain,

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

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

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

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

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

2. Originally Posted by ashura
Need help to find the particular integral $\displaystyle y_p$ of a 2nd order diffn eqn using Euler's eqn $\displaystyle only$;
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.

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

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

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

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

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

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

This is where i'm uncertain,

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

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

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

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

$\displaystyle \therefore y_p = Real \frac{e^{4ix}}{(4i-2)(4i-3)}] + 2\times Imaginary[\frac{e^{4ix}}{(4i-2)(4i-3)}]$ - Replace D with coefficient of x in $\displaystyle e^{4ix}$
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)
$\displaystyle (D^2-5D+6)y_p = 2sin4x$

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

Don't write your second step. The operator (D-2)(D-3) needs something to act on, so the $\displaystyle y_p$ 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 $\displaystyle [(D-2)(D-3)]^{-1}$ which we write in shorthand as $\displaystyle \frac{1}{(D-2)(D-3)}$.

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

-Dan

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

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

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

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

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

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

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

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

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

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

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

That's not the right answer which has cos4x in it. Where did I go wrong?
$\displaystyle y_p = 2 \mathbb{IM} \left ( \frac{e^{4ix}}{(4i - 2)(4i - 3)} \right ) = \frac{2}{25}cos(4x) - \frac{1}{25}sin(4x)$

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.

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

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

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

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

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

-Dan

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

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.

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

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

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

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

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

-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?: =

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

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

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

6. 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?: =

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

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

= $\displaystyle \mathbb{RE} \left ( cos(4x) \right ) + i \, 2 \mathbb{IM} \left ( sin(4x) \right ) \cdot \left ( - \frac{1}{50} + \frac{1}{25}i \right )$
Note that I didn't take the imaginary part until I had multiplied the two terms out. This is because:
$\displaystyle \mathbb{IM}(a + ib)(c + id) \neq (b)(d) = bd$

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

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

7. I think that I understand your point. I've worked the solution using cos4x + 2sin4x on the RHS. Thanks for your comment.

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

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

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

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

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

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

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

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

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

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

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

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

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