differential equations

#1

As soon as you can tell me how to solve the following:

y’’+4y’-4y = 0 for y in terms of only x

#2

Sounds like differential equations to me

#3

Yes, these types of equations tend to weed out certain folk.

#4

I haven’t taken DiffEq in years, but I do remember one thing in particular: guess and check!

Actually for that one, use eigenvectors. I don’t recall the exact details, but I’m pretty sure you can find the eigenvalues using a mathematical form of black magic I don’t precisely recall, then do… other stuff with them to solve it.

I swear I used to know this stuff!

EDIT: Okay, so the roots are -2±√(2). Which means the solution is… y = c1 * e^(-2+√(2)) + c2 * e^(-2-√(2)). Without initial values I can’t solve for c1 and c2.

BAM!

#5

Not only did you obtain the wrong roots, but your final expression is wrong; you’re missing a variable in the exponentials. So, let me save us all and force the devs to dedicate their lives to BMS =p.

First, assume that the solution is of the form y=Ae^(sx) (since y is a function of x in this case). Then, using this assumption, plug the function into the differential equation to obtain:

s^2+4s-4=0

Using the quadratic equation, you can then calculate the roots to obtain the following:

s = -2 +/- sqrt(32)/2

With the correct roots, you can define the function:

y = Ae^((-2 + sqrt(32)/2)x) + Be^((-2 - sqrt(32)/2)x)

Given the appropriate initial conditions, you can then solve for the constants.

#6

Actually, I just failed to type a 2 in mine – my work was correct on paper (the square root of 32 can be reduced to 4sqrt(2), which, when divided by 2, results in 2sqrt(2)).

But you are correct in that I forgot to list (x) in the exponents, and for that I apologize. My corrected answer would thus be

y = c1 * e^[(-2+2√(2))(x)] + c2 * e^[(-2-2√(2))(x)]

Which matches yours, so… I agree with your solution!

Told you I was rusty. shame
[COLOR=‘Black’]Sorry for going so far off topic, but I’m not sure there was a topic to begin with.

#7

So get working on the mod, Kane!

#8

And since C1 and C2 cannot be solved without initial conditions, which none are given, hence no solution can be found with y in terms of x only.

As far as I’m concerned, Quality over quantity. The less they update, the better the first impressions will be when this mod is released.

#9

:fffuuu:

#10

Bear with me and my lack of scripts :stuck_out_tongue:

Let y = e^ax
dy/dx = a * e^ax
d2y/dx2 = a^2 * e^ax

Therefore you have

a^2 * e^ax + 4a * e^ax - 4e^ax = 0

Divide by e^ax

a^2 + 4a - 4 = 0

Which is a nice quadratic equation. Solve to find a:

a_1 = 4.828 or a_2 = -0.828

Remember y = e^ax? Well we have two solutions, so

y_1 = e^(a_1)x and y_2 = e^(a_2)x

with our values for a:

y_1 = e^4.828x and y_2 = e^ -0.828x

Nearly there :stuck_out_tongue:

The general solution is given by y = Ae^(a_1)x + Be^(a_2)x
Where A and B are constants

So we have:

y = Ae^ 4.828x + Be^ -0.828x

If that’s wrong, well I’m pretty stuffed for my exam in 2 weeks :stuck_out_tongue:

EDIT: Ninja’d several times, but I’m proud of that and a couple of explanations can never hurt :slight_smile:

EDIT 2: I approve of maths threads :stuck_out_tongue: Do we need a separate thread for integration? :wink:

#11

Technically, you were the only one that actually SHOWED how to solve the problem, as originally requested by Kane. The rest of us just posted answers… so technically you win!

EDIT: And if you read this Catz, thanks for just segregating our inane side conversation, instead of mass-deleting. You rock!

#12

seven

#13

Your welcome Azrael. I am now waiting for the next problem to present itself for fan solving. James?

#14

I can follow some of it, but I get a little lost somewhere around the middle. I’m currently finishing up Calc 2. I’m squeezing 4 years of math into three + a microterm before attending another university to get my engineering studies in 2 years. With the degree plan I’m on I’ll graduate in 5 years (counting this one) from two universities, one with a bachelor’s in math and the other with and engineering degree.

#15

Sure,

y’’ + 4y = -5 * δ ( t - a )

where δ ( t - a ) satisfies

  1. δ ( t - a ) = 0 and t != a
  2. ∫ δ ( t - a ) dt = 1 ( integrate from -infinite to +infinite )

also y(0) = 1, y’(0)=0, a = 3

The techniques above are using the idea of Vector Space to solved DE.
https://tutorial.math.lamar.edu/Classes/LinAlg/VectorSpacesIntro.aspx

Its a quite intense, but its a hell of a lot easier to solve certain DE with it.
https://en.wikipedia.org/wiki/Linear_differential_equation#Homogeneous_equations_with_constant_coefficients

#16

Thanks for posting the link to the page that solves the problem in question :smiley: I learned this method around 4 years ago, so I wouldn’t remember by myself.

Ah! Dirac’s delta… “function”, good to see you again :slight_smile:

Haha! Here is the solution:

y(t) = (5/2) [sin(6) cos(2t) - cos(6) sin(2t)] deg(t - 3)

I tested and it works :smiley: Hurray!

(I struggled a bit to find it, at first I forgot a 2 somewhere and couldn’t find it).

I’ll leave for others to find it as well and maybe explain how it’s done.

EDIT:

deg is the degree function, i.e.

deg (x) = 1 if x >=0
deg (x) = 0 if x < 0

EDIT2:

Aw! Bad English! deg is not the degree function, it’s the step function.

#17

I got y(t) = (5/2)[sin( 6 - 2t ) )]*deg(t - 3) - cos( 2t )

you missed cos( 2t ), probably forgot something during the transformation ?

#18

ITT: wut

#19

:FFFUUU:

Ah!, dammit, you wrote y(0) = 1 and I saw y(0) = 0 :retard:

In this case, the solution should be

y(t) = (5/2) [sin(6) cos(2t) - cos(6) sin(2t)] deg(t - 3) + cos (2t) =
= (5/2) [sin(6) cos(-2t) + cos(6) sin(-2t)] deg(t - 3) + cos (2t) =
= (5/2) sin(6 - 2t) deg(t - 3) + cos(2t)

#20

Dammit, now I’m gonna have to go find my notes for that semester.

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