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Selfreferential integration 
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#1
Jun1003, 01:39 AM

P: 691

Suppose that we have a function that refers to itself in its derivative or second derivitive.
For example, let's say that you have a spring for which the force is directly proportional to the distance the spring has been compressed. F = cx For simplicity's sake, mass is constant, so we can just say a = cx The differentialform equation for this acceleration is: d^{2}x/(dt)^{2} = cx And so the speed equation is dx/dt = v_{0} + [inte]cx(dt) How would you solve this? Or, for a simpler equation, say that the velocity of a particle depends on its position: dx/dt = c_{1} + c_{2}x How would you solve this to get to the position function? 


#2
Jun1003, 05:50 AM

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PF Gold
P: 39,564

The exponential function is ideal for what you call "self referential" d (exp(x))/dx= x.
Since cos(x)= (exp(ix) exp(ix))/2 and sin(x)= (exp(ix) exp(ix))/(2i), sine and cosine also work. In particular if d^{2}x/dt^{2}= kx, (a "linear, homogeneous differential equation with constant coefficients") Then x(t)= C_{1}cos([sqrt](k)t)+ C_{2}sin([sqrt](k)t) where C_{1} and C_{2} can be any constants. For your second example, dx/dt= c_{1}+ c_{2}x, since this is a separable first order differential equation, we can write it in "differential form" and integrate: dx/dt= c_{1}+ c_{2}x is the same as dx/(c_{1}+ c_{2}x)= dt We can integrate the left by using the substitution u= c_{1}+ c_{2}x so du= c_{2}dx and the integral is (1/c_{2})[int](1/u)du= (1/c_{2})lnu or (1/c_{2})lnc_{1}+ c_{2}x. Integrating both sides, (1/c_{2})lnc_{1}+ c_{2}x= t+ C where C is the "constant of integration". Of course we can rewrite this as lnc_{1}+ c_{2}x= c_{2}(t+ C) and, taking the exponential of both sides, c_{1}+ c_{2}x= C' exp(c_{2}t) (C' is exp(c_{2}C) so that x is again an exponential function: x(t)= (1/c_{2})(C' exp(c_{2}t) c_{1}). 


#3
Jun1503, 11:39 AM

P: 691

You can't have x' = x + c for an a^{x} equation. You could have (x+c)' = x, but not x' = x + c 


#4
Jun1703, 07:10 AM

Math
Emeritus
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PF Gold
P: 39,564

Selfreferential integration
AN exponential function is any function that has the variable, x, as an exponent. THE exponential function is specifically e^{x}.
An exponential certainly will help with something like x'= x+ c. This is what is known as a "linear nonhomogeneous differential equation with constant coefficients". The simplest way to solve it is to drop the constant, c, so as to have a homogenous equation: x'= x whose solution is, of course, x(t)= Ae^{t} (A is an arbitrary constant). Since the differential equation is linear, we can simply add that to any solution of the entire equation. To get that constant c, note that the derivative of any constant is 0 so that if x= B (a constant) x'= 0 and the equation becomes 0= B+ c. x= B= c works. The general solution to the differential equation x'= x+ c is x(t)= Ae^{t} c. No, those are not the hyperbolic functions, those are the standard trigonometric functions. Since (sin t)'= cos t and (cos t)'=  sin t, they satisfy the equation x"= x. Since the hyperbolic functions can be written in terms of exponentials, they satisfy (cosh t)'= sinh t and (sinh t)'= (cosh t) and thus x"= x. 


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