Initial Condition?

2RIP

This question asks to solve the initial value problem and determine the interval in which it is defined.

$$y'=2x/(1+2y) , y(2)=0$$ and after solving, we get the equation $$y= -0.5 + \sqrt{x^{2} - 15/4}$$(discarded the negative squareroot since governed by initial condition).

Therefore, the interval is found to be $$x > or = \sqrt{15/4}$$, but my question is, why can't the interval also be defined at $$x < \sqrt{15/4}$$? How does it violate the initial condition? The I.C. only points out that when x=2, y=0.

I think i'm missing some basic knowledge for initial value problems...

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Defennder

Homework Helper
Is this restricted to the reals? If so, there's your explanation.

2RIP

Oh sorry, I was typing this question very late last night, i meant to mean $$x=-\sqrt{15/4}$$

Defennder

Homework Helper
Did you instead mean to say that why $$x \neq \pm \sqrt{15/4}$$ y' wouldn't be defined at that point, would it?

2RIP

Well I wasn't really considering whether y' played a role in defining the interval. I was just lost when the solution to that question was any x-value where $$x \geq + \sqrt{15/4}$$ and the negative of that wasn't part of the answer. I plugged in $$x=- \sqrt{15/4}$$ to y(x) and it gave me zero also. So why shouldn't $$x < -\sqrt{15/4}$$ be a solution as well?

HallsofIvy

Homework Helper
?? Well, again, if $x< -\sqrt{15/4}$, then $x^2> 15/4$ and your formula will involve a complex number.

Redbelly98

Staff Emeritus
Homework Helper
Maybe 2RIP means to ask: why isn't the interval

$$|x| \geq \sqrt{15/4}$$

?

Defennder

Homework Helper
Did you mean $$|x| \leq \sqrt{15/4}$$ instead?

Redbelly98

Staff Emeritus
Homework Helper
x2 needs to be greater or equal to 15/4 to avoid having a negative number under the radical.

$$y= -0.5 + \sqrt{x^{2} - 15/4}$$

HallsofIvy

Homework Helper
This question asks to solve the initial value problem and determine the interval in which it is defined.

$$y'=2x/(1+2y) , y(2)=0$$ and after solving, we get the equation $$y= -0.5 + \sqrt{x^{2} - 15/4}$$(discarded the negative squareroot since governed by initial condition).

Therefore, the interval is found to be $$x > or = \sqrt{15/4}$$, but my question is, why can't the interval also be defined at $$x < \sqrt{15/4}$$? How does it violate the initial condition? The I.C. only points out that when x=2, y=0.

I think i'm missing some basic knowledge for initial value problems...

Okay, $\sqrt{x^2- 15/4}$ is defined only for $x>\sqrt{15}/2$ or for $x< -\sqrt{15}/2$. You are asking why the solutions is not defined for $x< -\sqrt{15}/2$, right?

If you were to find the general solution to the differential equation it would involve a single undetermined constant, C say. The value of that constant can be determined by the initial value, y(2)= 0. Since $x= 2> \sqrt{15}/2$, that C applies to all $x>\sqrt{15}/2$. However, there is no way to "connect" values of the function for $x>\sqrt{15}/2$ and values of the function for $x< -\sqrt{15}/2$. There is no way of determining the "C" for $x< -\sqrt{15}/2$.

femme77

hi,
i have problem to determine initial condition for this problem. the system models a coupled oscillator
m1x''=-k1x+k2(y-x),
m2y''=-k2(y-x).
Imagine a spring (with constant spring k1) attached to a hook in the ceiling.Mass m1 is attached to the spring, and a second spring (with constant spring k2) is attached to the bottom of mass m1. by attaching mass m2 to the second spring, i have a coupled oscillator where x and y represent the displacements of masses m1 and m2 from their respective equilibrium positions. I have set x1-x, x2=x',x3=y and x4=y1 and i get
x1'=x2,
x2'=(-k1/m1)x1+(k2/m1)(x3-x1),
x3'=x4,
x4'=(-k2/m2)(x3-x1).
Assume k1=k2=1 and m1=m2=1.
Suppose that the first mass is displaced upward two units, the second mass downward two units, and both masses are released from rest, therefore the initial conditions for this problem are
x(0)=-2,x'(0)=0;y(0)=2;y',(0)=0;

Can anyone verify whether my stated initial conditions right or wrong... femme77

Does anyone can verify these initial conditions?

Redbelly98

Staff Emeritus
Homework Helper
Suppose that the first mass is displaced upward two units, the second mass downward two units, and both masses are released from rest, therefore the initial conditions for this problem are
x(0)=-2,x'(0)=0;y(0)=2;y',(0)=0;

Can anyone verify whether my stated initial conditions right or wrong... Yes, looks good, as long as you meant to use downward as the positive direction.

HallsofIvy

Homework Helper
hi,
i have problem to determine initial condition for this problem. the system models a coupled oscillator
m1x''=-k1x+k2(y-x),
m2y''=-k2(y-x).
Imagine a spring (with constant spring k1) attached to a hook in the ceiling.Mass m1 is attached to the spring, and a second spring (with constant spring k2) is attached to the bottom of mass m1. by attaching mass m2 to the second spring, i have a coupled oscillator where x and y represent the displacements of masses m1 and m2 from their respective equilibrium positions. I have set x1-x, x2=x',x3=y and x4=y1 and i get
x1'=x2,
x2'=(-k1/m1)x1+(k2/m1)(x3-x1),
x3'=x4,
x4'=(-k2/m2)(x3-x1).
Assume k1=k2=1 and m1=m2=1.
Suppose that the first mass is displaced upward two units, the second mass downward two units, and both masses are released from rest, therefore the initial conditions for this problem are
x(0)=-2,x'(0)=0;y(0)=2;y',(0)=0;

Can anyone verify whether my stated initial conditions right or wrong... Yes, those initial conditions are correct.

However, in the future, please do not "hijack" someone else's thread to ask a new question. It is very easy to start your own thread. Click on the "new topic" button at the top of the list of threads.

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