# Tough Differential Equation Problem

• May 2nd 2012, 07:00 PM
Calculon
Tough Differential Equation Problem
I need to find a function F such that it is continuous everywhere and

$\displaystyle y'(t)=F(y(t))$ and $\displaystyle y(0)=0$

The only thing i could think of is $\displaystyle y(t)=e^t$ but that obviously doesnt satisfy the initial value. Any help or hints is greatly appreciated.
• May 2nd 2012, 09:31 PM
MathoMan
Re: Tough Differential Equation Problem
So you're supposed to find both F and y(t) for which the given conditions are true, or ...?

How about $\displaystyle y(t)=\sin t$. Then $\displaystyle y(0)=\sin 0=0$.
And also $\displaystyle y'(t)=(\sin t)'=\cos t =\sqrt{1-\sin^2t}=\sqrt{1-(y(t))^2}=F(y(t)).$
• May 3rd 2012, 03:21 PM
Calculon
Re: Tough Differential Equation Problem
I forgot to mention that I need to find a function F such that the initial value problem has infinitely many solutions. I don't know if that changes anything....
• May 3rd 2012, 03:26 PM
Calculon
Re: Tough Differential Equation Problem
So to clear things up: I don't need to find a function y since this function F should take any function y and spit out its derivative i think.
• May 7th 2012, 08:19 AM
Ivanator27
Re: Tough Differential Equation Problem
The problem with this question is that the complexity of any derivative is arbitrary, for example, if we know that the function y is of the general form:

http://latex.codecogs.com/gif.latex?\large%20y(t)=at+b

... where a and b are constants. It is easy to create a function F that will reliably produce y's derivative, as in:

http://latex.codecogs.com/gif.latex?...frac{at}{at+b}

However, even a slight change in the general form of y will upset everything and F will no longer function, while, as stated earlier the above function F produces the derivative of y, it will not produce the derivative of x (with constants a, b and c):

http://latex.codecogs.com/gif.latex?\large%20y(t)=at+b
http://latex.codecogs.com/gif.latex?...0x(t)=(at+b)^c

http://latex.codecogs.com/gif.latex?...1-\frac{1}{c}}

However, a bit of manipulation will create a function that will work for both x and y, (above) function G, since they are quite similar. this won't work in all instances; consider the functions p and q, defined below:

http://latex.codecogs.com/gif.latex?...-t^{b}sin(ab)}
http://latex.codecogs.com/gif.latex?...frac{ab}{t}))}

Creating a function that will produce the derivatives of both these functions is an enormous, if not impossible, task. So, my conclusion is, it is possible to create a function F that will produce the derivative of a function y, but the general form of y must first be known. Otherwise we'd have to consider an infinite number of combinations of roots, fractions, logarithms, trigonometric functions and lord knows what else - which from my humble high-school perspective - seems impossible.

Sorry. =( I hope someone else has a more positive answer.