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C4 differential equation help needed

Hey. i cant seem to rearrange this equation correctly!

-15(dv/dt) = 2v - 450

When t= 0, v = 300
Find v, when t = 15

Thank you guys!

Reply 1

Revelation

-15(dv/dt) = 2v - 450
(dv/dt) = -(2/15)v + 30
v = -(2/15)vt + 30t + C

When t= 0, v = 300
300 = -(2/15)*300*0+30*0 + C
300 = C
C = 300

v = -(2/15)vt + 30t + C
v = -(2/15)vt + 30t + 300

Find v, when t = 15
v = -(2/15)15v + 30*15 + 300
v= -2v + 750
3v=750
v=250

Reply 2

~*tweety*~

ye thats the answer i got bt the book says 235 :s:

thanks 4 trying it out!! :smile:

Reply 3

Rabite

I get 235 (inexact).

-15(dv/dt) = 2v - 450
Separate variables: Div(2v-450) and div(-15) and Multi(dt)

1/(2v-450) dv = -1/15 dt
Int. both sides w.r.t. v and t.
½ln|2v-450| = -t/15 +c
Boundary condition:t= 0, v = 300
½ln|2(300)-450| = c
c = ½ln(150)
Put back into integrated equation
½ln|2v-450| = -t/15 + ½ln(150)
When t = 15:
½ln|2v-450| = -1 + ½ln(150) (Doubling gives...)
ln|2v-450| = -2 + ln(150) (Take the 150 thing to the other side)
ln|2v-450| - ln(150) = -2 (And combine logs)
ln(|2v-450| / 150) = -2
Eponentiate
e^ln(|2v-450| / 150) = e^-2
|2v-450| / 150 =e^-2
Solve for v
v = ½(150e^-2 + 450)

Erm. I think.

Reply 4

yusufu

Reply 5

~*tweety*~

ye thats rite its shud be approx 235, thank you sooo much!!

Reply 6

~*tweety*~

i have another query on differential equations :p:

Solve the d.e given that y = pi/4 at x = 0

e^x(dy/dx) = x/sin2y

thank you once again :biggrin:

Reply 7

Aired

Seperate the variables:

e^x.sin2y (dy/dx) = x
sin2y (dy/dx) = x/e^x
Integral: sin2y dy = Integral x.e^-x dx.

Then integrate both sides (the right hand side can be done by parts).

Reply 8

~*tweety*~

thanks, thats reli helpful!! :smile:

Reply 9

Rabite

In general:

Multiply through by d[whatever] so you get one d on one side, and one d on the other.
Them multiply and divide through by whatever until you have all of one variable on the left and all of one variable on the right.
You can't subtract variables to separate them (as far as I know...), because the 'd[thing]' has to apply to the whole of the side.

And remember the +c ^ ^;;

Reply 10

~*tweety*~

Rabite

You can't subtract variables to separate them (as far as I know...), because the 'd[thing]' has to apply to the whole of the side.

this is where i am going rong! rather than multiply/divide i was +/- :p:

Reply 11

yusufu

^^ Naughty!

Reply 12

~*tweety*~

YYYY

^^ Naughty!

:littleang :p:

im slowly getting the hang of them now!! :smile:

Reply 13

yusufu

Reply 14

~*tweety*~

dus ne1 have the C4 edexcel txtbook, coz im stuck on a question from there, but u need to see the graph! Mixed Ex 6L, Q2b :s:

Reply 15

azza777

in the first problem on this page dont u have 2 separate the variables then integrate.

Reply 16

Rabite

I can't remember straight off, and I don't know where my book is...
It's a parametric right? I think it was the one I got stuck on a while back ^^;;

Well, use (int)y dx = (int)y dx/dt dt

Find the 't' values of all of the critical points.
Integrate between the lowest bit and the origin to get the area of the triangle below the x-axis. It should be negative.
Then add on the little triangle bit that's also under the axis.

Then deal with the above-the-axis bit. Integrate from t=0 to whatever the top point is, then minus the area of the excess triangle.

Then add everything together.
My explanation skills are -5 on the one-to-ten scale, so don't blame yourself if I sound like I'm talking gibberish...