Solving an ODE: Seeking Suggestions

[metamathphys]

Hello I am trying to solve this ODE


dx/dt=(f(x)+g(t))^(1/2)


I have been recalling what I learn in my ODE course and looking at my old textbook but I did not find what method is appropiate to try...any suggestions?

Thank you very much!

 

[MathematicalPhysicist]

You can take another derivative wrt t to get:

d^2 x/dt^2 = [f'(x)dx/dt+g'(t)]/2(dx/dt) = f'(x)/2 + g'(t)dt/dx/2

Now try to solve d^2 x /dt^2 =f'(x)/2 and d^2 x / dt^2 =g'(t)/(2dx/dt)

Seperately, and then by superposition you get a solution for the ODE.
the second equation you can write the solution explicitly d(dx/dt)^2/dt = g'(t) by integration you get:
(dx/dt)^2 =g(t)+C
dx/dt = sqrt(g(t)+C) ...

I believe you get the picture now.

 

[the_wolfman]

You can take another derivative wrt t to get:

d^2 x/dt^2 = [f'(x)dx/dt+g'(t)]/2(dx/dt) = f'(x)/2 + g'(t)dt/dx/2

Now try to solve d^2 x /dt^2 =f'(x)/2 and d^2 x / dt^2 =g'(t)/(2dx/dt)

Seperately, and then by superposition you get a solution for the ODE.
the second equation you can write the solution explicitly d(dx/dt)^2/dt = g'(t) by integration you get:
(dx/dt)^2 =g(t)+C
dx/dt = sqrt(g(t)+C) ...

I believe you get the picture now.

Sorry, but this is wrong. The equation you get when you differentiate wrt t is
\frac{dx}{dt}\frac{d^2x}{dt^2}-\frac{1}{2}\frac{df\left(x\right)}{dx}\frac{dx}{dt}=\frac{1}{2}\frac{dg\left(t\right)}{dt}

The first term is nonlinear. The second term is also nonlinear, except for special cases of f(x). Thus this equation is nonlinear, and superposition does not hold!

 

[D H]

Sorry, but this is wrong. The equation you get when you differentiate wrt t is
\frac{dx}{dt}\frac{d^2x}{dt^2}-\frac{1}{2}\frac{df\left(x\right)}{dx}\frac{dx}{dt}=\frac{1}{2}\frac{dg\left(t\right)}{dt}

That too is wrong. A correct expression for the second derivative with respect to time is $2\ddot x(t) = f'(x) + g'(x)$

That said, I don't think that this will help in general.

Hello I am trying to solve this ODE

dx/dt=(f(x)+g(t))^(1/2)

What might help is solving for t as a function of x yielding $t=\int (f(x)+g(x))^{-1/2}\,dx$ and then finding the inverse function that yields x as a function of t.

 

[metamathphys]

First of all I wish to thank you all for taking some time looking at my post.

My first comment concerns all of those who suggest taking another time derivative. I do not agree with your results if I take another derivative what I get is:

\frac{d^2x}{dt^2}=\frac{f&#039;(x)\frac{dx}{dt}+\frac{dg}{dt}}{2\sqrt{f+g}}[\itex]<br /> <br /> Which does not look any easier...I do not know about any equation that simplifies when you perform higher derivatives. <br /> <br /> Regarding the comment of D H <br /> <br /> If I knew t as a function of x then I would be done no? All I had left to do is invert the relation to have x as a function of t and that all. So your method requires knowing the solution to the problem so I don&#039;t see how this can be of any help maybe I am misunderstanding something if it is so please show explicitly how to do it with some easy example

 

[the_wolfman]

That too is wrong. A correct expression for the second derivative with respect to time is $2\ddot x(t) = f'(x) + g'(x)$

Actually its not. In the original problem statement g is a function of t, not a function of x. If g was a function of x, then you would be correct.

First of all I wish to thank you all for taking some time looking at my post.

My first comment concerns all of those who suggest taking another time derivative. I do not agree with your results if I take another derivative what I get is:

\frac{d^2x}{dt^2}=\frac{f&#039;(x)\frac{dx}{dt}+\frac{dg}{dt}}{2\sqrt{f+g}}

Its the same thing. We are using the original equation (\frac{dx}{dt} = \sqrt{f(x)+g(t)}) to simplify the denominator:
\frac{d^2x}{dt^2}=\frac{f&#039;(x)\frac{dx}{dt}+\frac{dg}{dt}}{2\sqrt{f+g}}=\frac{f&#039;(x)\frac{dx}{dt}+\frac{dg}{dt}}{2\frac{dx}{dt}}


For the record, I recommend trying to directly solve the first order ode:

\frac{dx}{dt} = \sqrt{f(x)+g(t)}.


You might get lucky by taking more derivatives but I doubt it. In general first order nonlinear ODEs are the easiest to solve. Higher order non-linear ODEs get progressively harder to solve.

I also don't know if a general solution to this equation exists for arbitrary f and g. Do you know what f and g are.

 

[D H]

Actually its not. In the original problem statement g is a function of t, not a function of x. If g was a function of x, then you would be correct.

Stupid me. You are correct. I didn't read the OP properly.

I also don't know if a general solution to this equation exists for arbitrary f and g.

Of course not. There certainly must exist some function x(t) defined over some subinterval over which $f(x)$ and $g(t)$ are bounded and continuous. However, good luck finding that x(t) in general. There's no guarantee that $\frac{dx}{dt} = f(t)$ has a solution in the elementary functions, let alone $\frac{dx}{dt} = \sqrt{f(x)+g(t)}$.

Do you know what f and g are.

That is a key question.

 

[MathematicalPhysicist]

Yes, I believe you may be right, this trick of the derivative of square root seemed like it may work here.

So what's the context metamathphys?

 

[MathematicalPhysicist]

Actually its not. In the original problem statement g is a function of t, not a function of x. If g was a function of x, then you would be correct.



Its the same thing. We are using the original equation (\frac{dx}{dt} = \sqrt{f(x)+g(t)}) to simplify the denominator:
\frac{d^2x}{dt^2}=\frac{f&#039;(x)\frac{dx}{dt}+\frac{dg}{dt}}{2\sqrt{f+g}}=\frac{f&#039;(x)\frac{dx}{dt}+\frac{dg}{dt}}{2\frac{dx}{dt}}


For the record, I recommend trying to directly solve the first order ode:

\frac{dx}{dt} = \sqrt{f(x)+g(t)}.


You might get lucky by taking more derivatives but I doubt it. In general first order nonlinear ODEs are the easiest to solve. Higher order non-linear ODEs get progressively harder to solve.

I also don't know if a general solution to this equation exists for arbitrary f and g. Do you know what f and g are.

Well, I can look at it as an equation of energy which is not constant of time.

I.e dx/dt is the velocity and we have v^2-f(x)=g(t) (or Lagrangian depending on the plus minus sign). So it's actually an equation of an energy which is not necessarily constant with time.

How to solve this analytically, I am afraid I don't know, but I guess some special function is in order...