help to find an exact solution

robcru1 · **Joined:** Sat, 3 Dec 2011 13:31:06 UTC **Posts:** 11

I am trying to find an exact solution of x in this equation using seperation of variables
dx/dt=x^2/(t+1)
can anyone please help me
so far i have 1/x^2.dx=1/(t+1).dt
This where i get stuck
Please please please help
many thanks

Shadow · **Posted:** Sun, 4 Dec 2011 21:09:23 UTC

robcru1 wrote:

I am trying to find an exact solution of x in this equation using seperation of variables
dx/dt=x^2/(t+1)
can anyone please help me
so far i have 1/x^2.dx=1/(t+1).dt
This where i get stuck
Please please please help
many thanks

Umm. . . this is where you integrate. . . .

robcru1 · **Joined:** Sat, 3 Dec 2011 13:31:06 UTC **Posts:** 11

sorry i realise that is lnx^2=ln(t+1) correct or would i be best off moving x & (t+1) up to the top and have X^-2.dx=(t+1)^-1.dt and integrate from there?
This would give me X^-1/-1=(t+1) ?????

Shadow · **Posted:** Sun, 4 Dec 2011 21:24:23 UTC

robcru1 wrote:

sorry i realise that is lnx^2=ln(t+1) correct or would i be best off moving x & (t+1) up to the top and have X^-2.dx=(t+1)^-1.dt and integrate from there?
This would give me X^-1/-1=(t+1) ?????

The integral of ${1\over x^2}$ is definitely NOT $\log x^2$ , you should be able to see that by taking the derivative to check your work. The second version is correct, and again you can check by taking the derivative. And you have to integrate both sides, not just the one, remember, you have to do the same thing to both sides if you want things to balance, if you just integrate one side that's like adding 5 to just one side, you cannot expect things to be the same on both sides if you only do things to one of the sides. Also, don't forget your +C after integrating.

robcru1 · **Joined:** Sat, 3 Dec 2011 13:31:06 UTC **Posts:** 11

I have integrated both sides, surely the integral of (t+1)^1 is (t+1) or am i wrong which is more than likely.
If i get the intgration correct i pressume i then just need to transpose to just leave an x=? equation
I dont want to come across as a retard so please excuse my stupidity.

Shadow · **Posted:** Sun, 4 Dec 2011 21:37:30 UTC

robcru1 wrote:

I have integrated both sides, surely the integral of (t+1)^1 is (t+1) or am i wrong which is more than likely.
If i get the intgration correct i pressume i then just need to transpose to just leave an x=? equation
I dont want to come across as a retard so please excuse my stupidity.

$$\int {dt\over t+1}\ne t+1$ , how did you get that? Let u=t+1

and try again.

robcru1 · **Joined:** Sat, 3 Dec 2011 13:31:06 UTC **Posts:** 11

Not sure what you mean
If i integrate (t+1)-1 i raise the power by 1 and divide by the new power so what am i missing.
The help is appreciated

outermeasure · **Posted:** Sun, 4 Dec 2011 21:44:41 UTC

Locking topic. Duplicate.

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help to find an exact solution

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