# Second Order Differential Equations question

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#1
The angular displacement from its equilibrium position of a swing door is modelled by the differential equation:
(d2θ/dt2) + 4(dθ /dt) + 5θ = 0

Boundary condition : θ = π/4 when t = 0

How do I find the particular solution? Struggling with the constants.

Edit 1: Was thinking of differentiating to get velocity and make that 0 but not too sure
Last edited by Toast210; 8 months ago
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8 months ago
#2
Solve the differential equation, ignoring the boundary condition first. What have you tried?
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8 months ago
#3
(Original post by Toast210)
The angular displacement from its equilibrium position of a swing door is modelled by the differential equation:
(d2θ/dt2) + 4(dθ /dt) + 5θ = 0

Boundary condition : θ = π/4 when t = 0

How do I find the particular solution? Struggling with the constants.

Edit 1: Was thinking of differentiating to get velocity and make that 0 but not too sure
You realise the rhs is zero so the differential equation is homogeneous so the particular solution/integral is zero. Unless I've misunderstood something?
Last edited by mqb2766; 8 months ago
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#4
(Original post by mqb2766)
You realise the rhs is zero so the differential equation is homogeneous so the particular solution/integral is zero. Unless I've misunderstood something?
The particular solution and integral are different to my knowledge. I'm not sure the best way to describe it other than the particular solution is finding constants of the equation using the boundary conditions.
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#5
Solve the differential equation, ignoring the boundary condition first. What have you tried?
So I got the equation y = e^-2t(Acost + Bsint)

Using the first boundary condition, I got A as π/4 however I'm not sure about finding B
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8 months ago
#6
(Original post by Toast210)
The particular solution and integral are different to my knowledge. I'm not sure the best way to describe it other than the particular solution is finding constants of the equation using the boundary conditions.
In that case you really need two boundary values. You only have one so will have one degree of freedom remaining. You could assume d theta/dt is zero at t=0, but the question should clearly state that.
Last edited by mqb2766; 8 months ago
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#7
(Original post by mqb2766)
In that case you really need two boundary values. You only have one so will have one degree of freedom remaining. You could assume d theta/dt is zero at t=0, but the question should clearly state that.
See I was thinking that because displacement is 0, velocity is also 0. However, I wasn't too sure and wanted to clarify. But it seems that's the only way I could do it right?
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8 months ago
#8
(Original post by Toast210)
See I was thinking that because displacement is 0, velocity is also 0. However, I wasn't too sure and wanted to clarify. But it seems that's the only way I could do it right?
You can't necessarily assume that but it is a common occurrence with mechanical systems. With a 2nd order differential equation you need two initial / boundary conditions to solve it uniquely. The question should clearly state it. Where does it come from?
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8 months ago
#9
The door starts from rest so d theta/dt=0 when t=0.
Last edited by mqb2766; 8 months ago
0
8 months ago
#10
(Original post by Toast210)
See I was thinking that because displacement is 0, velocity is also 0. However, I wasn't too sure and wanted to clarify. But it seems that's the only way I could do it right?
Displacement isn't 0 - it's .

(Original post by mqb2766)
The door starts from rest so d theta/dt=0 when t=0.
Kind of depressing that skill in finding the actual question the OP has posted is so important on the forum these days, although I salute you for it! (PRSOM, sigh).
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8 months ago
#11
(Original post by DFranklin)
Kind of depressing that skill in finding the actual question the OP has posted is so important on the forum these days, although I salute you for it! (PRSOM, sigh).
I should have realised that the wording after the ode wasn't from the original question, but yes, it's always good to have a pic of the full question.
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8 months ago
#12
The solution has complex roots and these can be found by solving using the quadratic formula and substituting in using i=root(-1). Now you have two roots to the auxiliary equation and can form a general solution in the form theta= Acos(t)e^(-ct) Bsin(t)e^(-ct). Then you know theta(0)=pi/4, and theta'=0- so substituting in these values into the general solution and it's derivative the particular solution can be found. I get A=pi/4 and B=pi/2. Hope this helps 0
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