# Problem with a differential equation

• Apr 28th 2008, 06:55 PM
arbolis
Problem with a differential equation
I'm having some trouble solving a simple differential equation, and yes I'm a bit lazy to check my notes. (Nod)
It's a simple one, here it is : We have that $x'=f(t,x)$, $x(0)=0$. Use the fact that $dt/dx=(dx/dt)^{-1}$ when dx/dt≠0.
$f(t,x)=x^{-2}$.
It's certainly a beginner question but I have a doubt in this case : x'=dx/dt? I guess yes, since otherwise I don't see what could be x'!
• Apr 29th 2008, 12:39 AM
Opalg
Quote:

Originally Posted by arbolis
I'm having some trouble solving a simple differential equation, and yes I'm a bit lazy to check my notes. (Nod)
It's a simple one, here it is : We have that $x'=f(t,x)$, $x(0)=0$. Use the fact that $dt/dx=(dx/dt)^{-1}$ when dx/dt≠0.
$f(t,x)=x^{-2}$.
It's certainly a beginner question but I have a doubt in this case : x'=dx/dt? I guess yes, since otherwise I don't see what could be x'!

Yes, x' is an alternative notation for dx/dt.
• Apr 29th 2008, 01:21 PM
arbolis
Still in the mud
Thanks Opalg. This notation confuses me. I don't really understand well how it can be solved. Are we looking for a function x(t)? Or just "x"?
Also... well I need to read much more on calculus, but why can we separate the dt from the expression dx/dt? It's not a division...
I think I can solve problems like this one : $x'=t^3$ and with the same initial condition as I posted at first. But when I have x' in function of x, then I'm lost. (that's why I'm posting here right now).
It's also confusing to me that instead of writing x'(t)=t for example, we just write x'=t.
What I've thought about the problem is $\frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow dx=\frac{1}{x^2} dt$. Now from there should I integrate??? I will have x= something depending of x but with dt at the end. So I guess this is not the way to do it.
Can someone give me a hint solving the differential equation?
• Apr 29th 2008, 01:27 PM
icemanfan
Quote:

Originally Posted by arbolis
Thanks Opalg. This notation confuses me. I don't really understand well how it can be solved. Are we looking for a function x(t)? Or just "x"?
Also... well I need to read much more on calculus, but why can we separate the dt from the expression dx/dt? It's not a division...
I think I can solve problems like this one : $x'=t^3$ and with the same initial condition as I posted at first. But when I have x' in function of x, then I'm lost. (that's why I'm posting here right now).
It's also confusing to me that instead of writing x'(t)=t for example, we just write x'=t.
What I've thought about the problem is $\frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow dx=\frac{1}{x^2} dt$. Now from there should I integrate??? I will have x= something depending of x but with dt at the end. So I guess this is not the way to do it.
Can someone give me a hint solving the differential equation?

You're almost there. This differential equation is separable. Just take $\frac{dx}{dt} = \frac{1}{x^2}$ and multiply both sides of the equation by $x^2 dt$ to yield $x^2 dx = dt$. Now you can integrate both sides.
• Apr 29th 2008, 01:45 PM
Opalg
Quote:

Originally Posted by arbolis
This notation confuses me. I don't really understand well how it can be solved. Are we looking for a function x(t)? Or just "x"?

That's the trouble with the x' notation: it doesn't make explicit what the variable is. In this case, it is implicitly understood that x is a function of the variable t.

In the same way, if you see the notation y', it will usually mean dy/dx, but you have to look at the rest of the problem to see whether the independent variable is in fact x (or it may be t, or some other letter).
• Apr 29th 2008, 01:55 PM
arbolis
I think I made an error. Here is what I've done :
$\frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow \int x^2 dx=\int dt\Leftrightarrow x=(3(t-C))^{1/3}$ soon after this we realize that C which is the constant produced when integrating is equal to 0. So at the end I found that $x(t)=(3t)^{1/3}$. Is it right?
• Apr 29th 2008, 01:57 PM
icemanfan
Quote:

Originally Posted by arbolis
I think I made an error. Here is what I've done :
$\frac{dx}{dt}=\frac{1}{x^2}\Leftrightarrow \int x^2 dx=\int dt\Leftrightarrow x=(3(t-C))^{1/3}$ soon after this we realize that C which is the constant produced when integrating is equal to 0. So at the end I found that $x(t)=(3t)^{1/3}$. Is it right?

If you can assume that C is equal to zero, then that answer is correct.
• Apr 29th 2008, 02:11 PM
arbolis
Thanks both for your valuable help!