/sci/ - Science & Math

[Anonymous No.2350592]

since most of threads are troll threads as usual and I'm not interrupting anything important I will kindly ask you to help me with some basic calculus shit ...

I need to know how to solve

y' + 2y = 0 y(0)=3

I always hit a dead end -.-

so plz help me learn something new

[Anonymous No.2350608]

>>2350592
>y' + 2y = 0 y(0)=3
>0 y(0)=3
>0=3

wat

[Anonymous No.2350607]

bump for dif equation help

[Anonymous No.2350615]

>>2350608
gb2precalc

[Anonymous No.2350630]

it is initial value
y(0)=x

[Anonymous No.2350638]

>>2350615
>gb2algebra
fixt.

[helpfag !!2GURbiVOGev No.2350642]

>>2350592
integrate the shit on the left

you have an equation with an integrating constant, and point on that equation which was already given to you.

plug that point in and solve for the integrating constant

if i do it for you, you'll come back asking for help again and we can't let that happen

[Anonymous No.2350675]

man I tried I fuck something up -.-

I really think I just need to see one done like that ... I get some weird shit that is wrong ..

I dont give a fuck about solution , but if you can do something similar It will be fine ...

I just want to know how its done ..but book I'm learning from doesn't have that kind of examples -.-

weird shit is I've done like 10-15 similar but with anything but 0 on right side...

If you won't than tnx anyways for what you wrote so far ...

[Anonymous No.2350710]

<div class="math">
y' + 2y = 0</div><div class="math">
y' = -2y </div>
y is its own derivative times -2. What function is its own derivative? Of course:
<span class="math">
y = ke^{-2x}
[/spoiler] and since <span class="math"> y(0) = 3 [/spoiler], <span class="math"> y = 3e^{-2x} [/spoiler]

[Anonymous No.2350722]

>>2350710


tnx bro highly appreciated

[Anonymous No.2350852]

>>2350710

w8 isn't initial value saying that y=3 for x=0

and then I find solution y=x+c ? and that will give me constant C ?

kinda confused again :)

[Anonymous No.2350896]

<span class="math">\mathcal{L}(y'+2y)=\mathcal{L}(0)[/spoiler]
<span class="math">s\mathcal{L}(y)-y'(0)+2\mathcal{L}(y)=0[/spoiler]
<span class="math">(s+2)\mathcal{L}(y)=3[/spoiler]
<span class="math">y=3\mathcal{L}^{-1}(\frac{1}{s+2})[/spoiler]
<span class="math">y=3e^{-2t}[/spoiler]

[Anonymous No.2350928]

If you are stuck you can always resort to the laplace transform:

<span class="math">\mathcal{L}(y'+2y)=\mathcal{L}(0)[/spoiler]
<span class="math">s\mathcal{L}(y)-y'(0)+2\mathcal{L}(y)=0[/spoiler]
<span class="math">(s+2)\mathcal{L}(y)=3[/spoiler]
<span class="math">y=3\mathcal{L}^{-1}(\frac{1}{s+2})[/spoiler]
<span class="math">y=3e^{-2t}[/spoiler]

[Anonymous No.2350953]

>>2350928

>as if he'd know how to do that

[Anonymous No.2350988]

>>2350928

Not OP but I'd love to know how that shit works. I guess there's no brief explanation to tide me over till I can go and study it?

[Anonymous No.2351010]

>>2350988
Further into your textbook, if you are studying from one. Laplace transforms make many differential equations easy.

[Anonymous No.2351033]

>>2350988
http://www.khanacademy.org/video/laplace-transform-1?playlist=Differential%20Equations

[Anonymous No.2351039]

If you wanted to work it all out you'd use seperation of variables.

dy/dt + 2y = 0
dy/dt = -2y
multiply the dt over to the right, and divide by the y
1/y dy = -2 dt
Now integrate both sides
ln(y) = -2t + c
Raise each side to the power e
y = e^(-2t + c)
y = e^(-2t)*e^c
e^c is just another constant, so you now you have
y = ce^(-2t)
Now apply initial conditions.
3 = ce^0
c = 3
So your final answer is
y = 3e^(-2t)

[Chooch51 !!FnpSzof2GZy No.2351048]

>>2350592
no

[Anonymous No.2351049]

>>2350988

The "shortcut" way is to first solve for y'= - 2y and then apply a technique called seperation of variables. This is easier seen by writing y' as the quotient of dy/dx

dy/dx= - 2y

Then treat dy and dx not as a single combined entity, but as seperable quantities. The goal is to get dy on the same side as all y's, and dx on the same side of all x's. This isn't always possible for all differential equations, but is for this one. Just multiply by dx, and divide by y

(1/y) dy = -2 dx

Then integrate normally. You would normally have a constant on both sides, but when you isolate y, you move the two constants together, so it is easiest to just make one constant (since a constant minus a constant is a constant, this is valid. Just keep the constants straight).

(1/y) dy = -2 dx
ln lyl = -2x +c
y= e^ (-2x+c)
y=c e^(-2x)

e raised to a constant is a constant, and e^(a+b) can of course be rewritten as e^a e^b, so that is where the last step comes from. In terms of being concise, using the same variable c as the constant really isn't proper, but you eventually get a foresight for what you're ultimate constant will be. Until then, just use subscripts (ie. C1 C2 C3, etc.)

So now we take our general solution y= c e^(-2x) and our initial condition y(0)=3 and solve for c

3= c e^0 = c
c=3

So the specific solution is

y=3 e^(-2x)

[Anonymous No.2351147]

>>2350928
Not OP but you got me interested Laplace transforms. I think I've got it all, but how do I take the inverse Laplace transform of a function?
I'm using y' = -1, and I've reached y = L^-1(-1/s^2) which should equal -t.

[Anonymous No.2351218]

>>2351147
Go be dumb somewhere else

[Anonymous No.2351220]

>>2351147

It's pretty much just a matter of working backwards and making things match (use the table I just posted). If you're using second order equations, you're usually going to have to do some partial fraction expansion.

[Anonymous No.2351225]

>>2351220
Thanks