2nd Order Differential Equation + Boundary Conditions!

Vespre · **Joined:** Thu, 8 Mar 2012 14:46:00 UTC **Posts:** 4

Hi guys, I have been doing some questions on complex numbers and second order differential equations. It's been going mostly fine, but now I have hit a bit of brick wall in my mind.

The question is this:

Solve the equation:

$\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2}+4\dfrac{\mathrm{d}y}{\mathrm{d}x}+4y=\cos(x)$

subject to the boundary conditions y(0) = 0 and y(pi) = 0

I can solve it fine, I get: Ae^-2x + Bxe^-2x + 4/25sinx + 3/25cosx

But I can't get my head around what I am supposed to do with the boundary conditions, I'm completely lost :confused:

Does anybody have any ideas? Any help would be great thanks

Shadow · **Posted:** Thu, 8 Mar 2012 16:40:10 UTC

Vespre wrote:

Hi guys, I have been doing some questions on complex numbers and second order differential equations. It's been going mostly fine, but now I have hit a bit of brick wall in my mind.

The question is this:

Solve the equation:

subject to the boundary conditions y(0) = 0 and y(pi) = 0

I can solve it fine, I get: Ae^-2x + Bxe^-2x + 4/25sinx + 3/25cosx

But I can't get my head around what I am supposed to do with the boundary conditions, I'm completely lost :confused:

Does anybody have any ideas? Any help would be great thanks

Use the boundary conditions to solve for A and B!

Vespre · **Joined:** Thu, 8 Mar 2012 14:46:00 UTC **Posts:** 4

I got that bit

I'm just not sure how to. Do I sub 0 into the equation somehow? What about the pi one?

If anybody can show me or at least show me a little bit to help get me started, that would be great.

Thanks for the quick reply though

Shadow · **Posted:** Thu, 8 Mar 2012 16:54:40 UTC

Vespre wrote:

I got that bit

I'm just not sure how to. Do I sub 0 into the equation somehow? What about the pi one?

If anybody can show me or at least show me a little bit to help get me started, that would be great.

Thanks for the quick reply though

That's certainly what y(0)

and $y(\pi)$ mean, right? That's not differential equations, it's definition of how functions work.

Vespre · **Joined:** Thu, 8 Mar 2012 14:46:00 UTC **Posts:** 4

Well the equation itself is a second order differential, so you solve it with the axillary equation (set it to = 0), then solve the particular integral (right hand side bit) to get the general solution.

It then just asks me to do solve it 'subject to the boundary conditions y(0) = 0 and y(pi) = 0. So I assume it just means set my solved equation to 0 and stick 0 where x is? And for the other one, do the same, but stick pi where x is? And it should give me some values for A and B that I can stick in to give me a fully solved equation?

Sorry if I completely missed your point there, I think I've confused myself more than I needed to

Shadow · **Posted:** Thu, 8 Mar 2012 21:00:34 UTC

Vespre wrote:

Well the equation itself is a second order differential, so you solve it with the axillary equation (set it to = 0), then solve the particular integral (right hand side bit) to get the general solution.

It then just asks me to do solve it 'subject to the boundary conditions y(0) = 0 and y(pi) = 0. So I assume it just means set my solved equation to 0 and stick 0 where x is? And for the other one, do the same, but stick pi where x is? And it should give me some values for A and B that I can stick in to give me a fully solved equation?

Sorry if I completely missed your point there, I think I've confused myself more than I needed to

He already did those things, he just didn't realize to use the boundary conditions.

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2nd Order Differential Equation + Boundary Conditions!

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