Kedas Posted April 26, 2006 Share Posted April 26, 2006 (Dy)² = (a-y²) with y(0) = 0 How do you solve this? Matlab gives these solutions: syntax: dsolve('(Dy)^2 = (a-y^2)','y(0) = 0') y=tan(t)*(a/(tan(t)^2+1))^(1/2) and y=-tan(t)*(a/(tan(t)^2+1))^(1/2) But these solutions are periodic if you look at the diff. equation: when y²=a then y should stay constant (not periodic) can anyone help? Link to comment Share on other sites More sharing options...

Perturbation Posted April 26, 2006 Share Posted April 26, 2006 It's seperable [math]\frac{dy}{dx}=\sqrt{a-y^2}[/math] [math]\frac{1}{\sqrt{a-y^2}}\frac{dy}{dx}=1[/math] [math]\int\frac{1}{\sqrt{a-y^2}}\frac{dy}{dx}dx=\int\frac{1}{\sqrt{a-y^2}}dy=\int dx[/math] Link to comment Share on other sites More sharing options...

Kedas Posted April 26, 2006 Author Share Posted April 26, 2006 let's make it [math]\frac{dy}{dt}=\sqrt{a^2-y^2}[/math] [math]\int\frac{1}{\sqrt{a^2-y^2}}dy=\int dt = sin^-1 \frac{y}{a} = t[/math] so [math]y=a.sin(t)[/math] But this is still periodic, why is y not staying constant when it reaches a? Link to comment Share on other sites More sharing options...

s pepperchin Posted April 26, 2006 Share Posted April 26, 2006 You are looking at the problem wrong. The answer says that when t is a half integer multiple of Pi, such as Pi times 1/2, 3/2, 5/2 etc, then x is equal to a, just like the graph shows. Your solution is a function of time not a. Link to comment Share on other sites More sharing options...

nicobudini Posted April 26, 2006 Share Posted April 26, 2006 (Dy)^2 stands for [math](\frac{dy}{dx})^2[/math] or [math]\frac{d^2y}{dx^2}[/math]? Link to comment Share on other sites More sharing options...

Kedas Posted April 27, 2006 Author Share Posted April 27, 2006 (Dy)^2 stands for [math](\frac{dy}{dx})^2[/math] or [math]\frac{d^2y}{dx^2}[/math']? The first one Link to comment Share on other sites More sharing options...

Kedas Posted April 27, 2006 Author Share Posted April 27, 2006 You are looking at the problem wrong. The answer says that when t is a half integer multiple of Pi' date=' such as Pi times 1/2, 3/2, 5/2 etc, then x is equal to a,just like the graph shows. Your solution is a function of time not a.[/quote'] The problem is the same as loading a capacitor only with a kwadratic equation. a is the target value and y is the rising voltage. dy/dt represents the current. (or speed in which the voltage changes.) Link to comment Share on other sites More sharing options...

Kedas Posted April 28, 2006 Author Share Posted April 28, 2006 If you solve this (Dy) = (a²-y²) with y(0) = 0 then y=a for t=inf (not periodic) y = a * (exp(2*t*a)-1) / (exp(2*t*a)+1) y= a * (F -1) / (F + 1) that sqrt is making it periodic somehow... Link to comment Share on other sites More sharing options...

Yggdrasil Posted April 28, 2006 Share Posted April 28, 2006 When you solve a differential equation, it is not guaranteed that your soultion will exist for all time. Note that dy/dx is undefined when y > sqrt(a). Therefore, when your trajectory passes y = sqrt(a), your solution ceases to exist. In the example you provided in the opening post, the solution you plot is valid from 0<t<pi/2. When you apply this restiction to the domain, the soultion is not periodic (as any soultion to a 1st order ODE should be). Link to comment Share on other sites More sharing options...

Kedas Posted April 28, 2006 Author Share Posted April 28, 2006 When you solve a differential equation, it is not guaranteed that your soultion will exist for all time. Note that dy/dx is undefined when y > sqrt(a). Therefore, when your trajectory passes y = sqrt(a), your solution ceases to exist. In the example you provided in the opening post, the solution you plot is valid from 0<t<pi/2. When you apply this restiction to the domain, the soultion is not periodic (as any soultion to a 1st order ODE should be). (I will talk about the second equation with a² because that's easier) Yes, I forgot about that part where y isn't defined. but there is one other thing I don't see: Dy is defined for y=a (Dy=0) but y should never get to that point 'a' since Dy reaches zero when y reaches that point. Just like the diff. eq. without sqrt, y will never reach 'a'. Link to comment Share on other sites More sharing options...

woelen Posted April 28, 2006 Share Posted April 28, 2006 Look at it in another way. As long as |y|< a, then dy/dt > 0. So, y can only increase. Given the initial value y(0) = 0. The solution moves towards a as time passes on, but at t = pi/2, the solution reaches y = a, and from that point it remains there. So, the solution is as follows: y(t) = a*sin(t) for 0<= t <= pi/2. y(t) = a for t > pi/2 This is a continuous function of time. Just like the diff. eq. without sqrt, y will never reach 'a'. This statement is not true. The solution does not move asymptotically to a, as opposed to the solution of dy/dt = a - y, it actually reaches a at time pi/2. Look at the explanation below. When y is very close to a (let's say y = a - e, e close to zero, but positive), then dy/dt = -de/dt = sqrt(a*a - a*a + 2*a*e - e*e) because e is very small, the following approximation is valid: -de/dt = sqrt(2a) * sqrt(e) or: de/dt = -sqrt(2a)*sqrt(e) de/sqrt(e) = -sqrt(2a)*dt ==> 2sqrt(e) = -sqrt(2a)*t + C, with C a positive constant. You see that sqrt(e) goes to zero linearly with time and e itself goes to zero quadratically. Hence, when y is near a, then y will reach a quadratically, and that is exactly what is the case near the top of the sine curve. Once, this point is reached, y will stick there forever and the system becomes stationary. Link to comment Share on other sites More sharing options...

Kedas Posted April 28, 2006 Author Share Posted April 28, 2006 Thanks, Then my math model or my assumption of what will happen is the problem. It's about filling a tank with air. (y=tank pressure, a=inlet pressure. (dy/dt)=airflow) Does this mean that at a certain point the pressure in the tank will reach the applied pressure or is my airflow formula not good enough? (it seems logical to assume that it will never reach that pressure like a capacitor) formula in detail http://www.scienceforums.net/forums/showthread.php?t=20182 Link to comment Share on other sites More sharing options...

woelen Posted April 28, 2006 Share Posted April 28, 2006 If this differential equation is your model, then I can say immediately that it is not a good model. Indeed, the pressure will go to its final value asymptotically. This differential equation does not, so it cannot be a good description. Link to comment Share on other sites More sharing options...

Kedas Posted April 28, 2006 Author Share Posted April 28, 2006 If this differential equation is your model, then I can say immediately that it is not a good model. Indeed, the pressure will go to its final value asymptotically. This differential equation does not, so it cannot be a good description. The model is based on this: (bottom page) http://www.engineersedge.com/fluid_flow/pressure_drop/pressure_drop.htm and also used here: http://www.cpvmfg.com/cat98/g-series15.htm Link to comment Share on other sites More sharing options...

woelen Posted April 28, 2006 Share Posted April 28, 2006 I've been thinking about the formula at the second link. This formula for pressure drop seems to be for a stationary situation, where there is a constant gas flow. I'm not an expert on this type of systems, but I have the idea that you cannot easily derive an equation from this for the type of problem you want to solve, with a step-shaped pressure at the input. Link to comment Share on other sites More sharing options...

Perturbation Posted April 28, 2006 Share Posted April 28, 2006 let's make it[math]\frac{dy}{dt}=\sqrt{a^2-y^2}[/math] [math]\int\frac{1}{\sqrt{a^2-y^2}}dy=\int dt = sin^-1 \frac{y}{a} = t[/math] so [math]y=a.sin(t)[/math] But this is still periodic' date=' why is y not staying constant when it reaches a?[/quote'] That should be root a by the way in the argument of the inverse sine, as [math]\frac{d(arcsin\tfrac{x}{a})}{dx}=\frac{1/a}{\sqrt{1-x^2/a^2}}=\frac{1}{\sqrt{a^2-x^2}}[/math] Whatever it is that you've used to solve and graph it is just extending the range of the sine function to be the reals. Your y(t) is only defined on the range [-90, 90]. Link to comment Share on other sites More sharing options...

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