# few questions with laplace transform

• May 7th 2011, 07:39 PM
Jeonsah
few questions with laplace transform
I am having trouble this this problem.
$\displaystyle y' + 6y = e^4^t, y(0) = 2$

Heres the work I have up until where I get stuck:
$\displaystyle SY(s) - y(0) + 6Y(s) = \frac{1}{s-4}$
$\displaystyle SY(s) - 2 + 6Y(s) = \frac{1}{s-4}$
$\displaystyle SY(s) + 6Y(s) = \frac{1}{s-4} + 2$
$\displaystyle Y(s)(s+6) = \frac{1}{s-4} + 2$
$\displaystyle Y(s) = \frac{1}{(s-4)(s+6)} + \frac{2}{s+6}$
partial fractions:
$\displaystyle \frac{1}{(s-4)(s+6)} + \frac{2}{s+6} = \frac{A}{s-4} + \frac{B}{s+6}$

How do I solve this partial fraction equation now? I have tried it and I am ending up with a different answer than the solutions manual for B.

My other question is:
Find the fundeamental set of solutions using only real coefficients: $\displaystyle y^(^4^) + 4y = 0$

I get the steps:
$\displaystyle m^4 + 4 = 0$
$\displaystyle m = \sqrt[4]{-4}$
$\displaystyle y1 = e^xcosx$
$\displaystyle y2 = e^xsinx$
$\displaystyle y3 = e^-^xcosx$
$\displaystyle y4 = e^-^xsinx$

I am lost as to where you get $\displaystyle m = \sqrt[4]{-4}$ is equal to $\displaystyle \pm 1 \pm i$ and how this works on a graph....

Thanks for the help guys, I really appreciate it!
• May 7th 2011, 07:44 PM
TKHunny
You cause me pain. How can you even SAY "laplace transform" and still be struggling with this basic algebra?

I get A = 1/10 and B = 19/10. What do you get and how?
• May 7th 2011, 08:08 PM
topsquark
Quote:

Originally Posted by Jeonsah
My other question is:
Find the fundeamental set of solutions using only real coefficients: $\displaystyle y^(^4^) + 4y = 0$

I get the steps:
$\displaystyle m^4 + 4 = 0$
$\displaystyle m = \sqrt[4]{-4}$
$\displaystyle y1 = e^xcosx$
$\displaystyle y2 = e^xsinx$
$\displaystyle y3 = e^-^xcosx$
$\displaystyle y4 = e^-^xsinx$

I am lost as to where you get $\displaystyle m = \sqrt[4]{-4}$ is equal to $\displaystyle \pm 1 \pm i$ and how this works on a graph....

Thanks for the help guys, I really appreciate it!

Your four solutions for m are wrong.

$\displaystyle m^4 + 4 = 0$

$\displaystyle m = (-4)^{1/4} = \sqrt{2} \cdot (-1)^{1/4}$

Recall that $\displaystyle -1 = e^{i \pi} = e^{i \pi + i2n \pi}$

So
$\displaystyle m = (-4)^{1/4} = \sqrt{2} \cdot (e^{i \pi + i2n \pi})^{1/4} = \sqrt{2} \cdot e^{i \pi / 4 + in \pi /2}$

$\displaystyle m = \sqrt{2} ( cos( \pi / 4 + n \pi / 2) + i~sin( \pi / 4 + n \pi / 2) )$

So the four solutions to m are, using n = 0, 1, 2, and 3:
$\displaystyle \{ cos(\pi / 4) + i~sin(\pi / 4),~cos(3 \pi / 4) + i~sin(3\pi / 4),~cos(5 \pi / 4) + i~sin(5\pi / 4),~cos(7 \pi / 4) + i~sin(7\pi / 4) \}$

I'll let you make the solutions look nicer.

-Dan

Edit: Whoops! I forgot about the factor of sqrt(2). Okay, so the final (simplified) solutions for m will be 1 + i, -1 + i, -1 - i, and 1 - i. Notice that this does generate the solution series given by your professor.
• May 7th 2011, 08:09 PM
Jeonsah
The reason why i was lost with the partial fraction part was because of the $\displaystyle \frac{2}{s+6}$ Exactly what do you do with it? do you combine it with the term on the left of the equal side, or do you move it over?
• May 7th 2011, 08:13 PM
topsquark
Quote:

Originally Posted by Jeonsah
The reason why i was lost with the partial fraction part was because of the $\displaystyle \frac{2}{s+6}$ Exactly what do you do with it? do you combine it with the term on the left of the equal side, or do you move it over?

You get a common denominator and you add it.
$\displaystyle \frac{1}{(s - 4)(s + 6)} + \frac{2}{s + 6} = \frac{1}{(s - 4)(s + 6)} + \frac{2(s - 4)}{(s - 4)(s + 6)}$

Finish adding the fractions, then use partial fractions on it.

-Dan
• May 7th 2011, 08:15 PM
Jeonsah
ah ok, thats where I got tripped up. I was adding it wrong. Thanks for helping me solve the second problem. That makes a lot more sense than how my teacher did it. Gracias Senor!
• May 7th 2011, 08:19 PM
Jeonsah
TKHunny, sorry for being a noob. Sometimes you just forget somethings you know? Ive been studying so much that im getting delirious. But none the less {shakes head} = me slapping myself!
• May 8th 2011, 04:08 AM
TKHunny
I hope you don't mind my friendly ribbing. If I received a paper like this during a class, I would mark it, "I should fail you for the entire semester for this".

One must remember ones algebra! :-)
• May 8th 2011, 04:34 AM
Prove It
Quote:

Originally Posted by TKHunny
I hope you don't mind my friendly ribbing. If I received a paper like this during a class, I would mark it, "I should fail you for the entire semester for this".

One must remember ones algebra! :-)

And if I was a head of a department, I would write a note saying "I should fire you for writing such a demotivational statement on a student's work..."
• May 8th 2011, 09:44 AM
TKHunny
Quote:

Originally Posted by Prove It
And if I was a head of a department, I would write a note saying "I should fire you for writing such a demotivational statement on a student's work..."

Ah, but if you were in my class, or if you were my department head, you would never take it as demotivational. You just can't see the huge smile on my face, the care and compassion in my eyes, and the personal commitment to make sure each student who cares is allowed to succeed and enabled to do so.
• May 8th 2011, 09:13 PM
Prove It
Quote:

Originally Posted by TKHunny
Ah, but if you were in my class, or if you were my department head, you would never take it as demotivational. You just can't see the huge smile on my face, the care and compassion in my eyes, and the personal commitment to make sure each student who cares is allowed to succeed and enabled to do so.

That is true, but the department head or principal wouldn't be able to see that either - any complaints and the only tangible evidence is what's written on the paper...