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Differential Equations help?

Hi, I think I've either gone wrong somewhere throughout this, or I need help establishing how to get to the answer for the second part, because I don't see how I would do it :/

Use integration to solve the differential equation dt/dx = (1/40)(x-15) where x>15, given that x=85 when t=0, expressing t in terms of x.

My working:
dt = (1/40)(x-15).dx
40.dt = (x-15).dx
integrate both sides
40t = (x²/2)-15x+c

I then subbed in x=85, t=0

0=(85²/2)-15(85)+c
0=3612.5-1275+c
0=2337.5+c
c= -2337.5

Making my equation 40t=(x²/2)-15x-2337.5

The second part of the question is "Hence show that x=15+70e^-t/40

How would I do this? :s-smilie:

Reply 1

Tottenh

Works if its dx/dt instead...although im not getting a minus on the t

Reply 2

SpringNicht

Original post by Tottenh

Works if its dx/dt instead...although im not getting a minus on the t

Oops, just looked at the question and it is dx/dt! :colondollar:

40.1/dt = (x-15).1/dx
1/40.dt = 1/(x-15).dx
1/(40t) = ln(x-15)+c
40t = 1/(ln(x-15)+c)
0=1/(ln60+c)
e⁰=e^ln60+c
1=60+c
59=c

I still don't think this is right though, because when I try and work it through I get this

40t=1/(ln(x-15)+59)
t=1/40(ln(x-15)+59)
1/t=40(ln(x-15)+59)
t^-1/40=ln(x-15)+59
e^{t^-1/40}=x-15+e⁵⁹
15+e^{t^-1/40}-e⁵⁹=x

:confused:

Reply 3

Tottenh

85-15 is 70.
So subbing in your conditions as soon as you have integrated gives:
ln70 = 0 + c, making c ln 70

Keep me updated

Reply 4

Tottenh

Also, when you do something to each side of the equation, you do it to ALL of it. You aren't raising the c to the power e.
But you also don't raise ln 70 to the power of e and then c, you raise them both TOGETHER.

i.e:
e^(ln60+c) which is defined as e^ln60 MULTIPLIED by e^c by the law of logs.

^ Obviously i am using your incorrect working as an example their but its just to show.

Reply 5

Killjoy-

Is the differential equation

\frac{dx}{dt}=\frac{(x-15)}{40}

Reply 6

SpringNicht

Original post by Tottenh

85-15 is 70.
So subbing in your conditions as soon as you have integrated gives:
ln70 = 0 + c, making c ln 70

Keep me updated

Oh dear

What am I doing at all.

Laws of logs - ln(x-15) - ln70 is the same thing as ln((x-15)/70) right?

e^t/40=x-15/70
70e^t/40=x-15
15+70e^t/40=x

I'm thinking my teacher missed typing in a minus sign on the 1/40 initially, as it works if I stick that in.

Making the question dx/dt=-(1/40)(x-15) with the same constraints

1/(x-15).dx = -1/40.dt
ln(x-15) = -t/40 +c
ln70 = 0/40 +c
ln70 = c
e^-t/40=x-15/70
70e^-t/40=x-15
15+70e^-t/40=x :biggrin:

Thankyou very much!

Reply 7

Tottenh

No problem man. Nicely done.

Reply 8

arvin_infinity

Original post by SpringNicht

Oh dear

Thankyou very much!

Just following the steps
when I multiply both sides by 40
I get
40ln|70|=0+C
therefore c=40ln|70|
Maybe I've gone crazy :colondollar:

Reply 9

SpringNicht

Original post by arvin_infinity

Just following the steps
when I multiply both sides by 40
I get
40ln|70|=0+C
therefore c=40ln|70|
Maybe I've gone crazy :colondollar:

You've forgot to multiply the constant, you'd have 40ln70 = 0 +40c :smile:

Reply 10

arvin_infinity

Original post by SpringNicht

You've forgot to multiply the constant, you'd have 40ln70 = 0 +40c :smile:

+rep
Oh it makes sense now!! that was the problem I did not know when to add the C ..I mean I multiplied it by 40 then I add the C (if you see what I mean)

Reply 11

SpringNicht

Original post by arvin_infinity

+rep
Oh it makes sense now!! that was the problem I did not know when to add the C ..I mean I multiplied it by 40 then I add the C (if you see what I mean)