Mixing
Using separation of variables to solve the wave equation
$begingroup$
I'm trying to teach myself separation of variables and have been following some notes for the wave equation, but there's one part which really confuses me and I'm not exactly sure how it makes the step.
For the wave equation
$$u_{tt} - c^2 u_{xx} = 0$$
with length $l$ and fixed ends, $u(0,t)=u(l,t)=0$ we seek a solution in the form $$u(x,t)=v(x)q(t)$$
and substituting this into the equation gives
$$frac{1}{q(t)}frac{d^2 q}{dt^2}=c^2 frac{1}{v(x)}frac{d^2 v}{dx^2}=-omega ^2$$
and I understand all of this so far. But when solving the equation
$$q''+omega ^2 q = 0$$
the general solution is
$$q(t)=A cos(omega t +alpha)$$
where $A$ and $alpha$ are constants - which is the part which I'm not exactly sure on how it gets to.
Could someone explain this step to me? Thanks!
pde wave-equation
edited Jan 31 at 17:42
Adrian Keister
5,26971933
asked Jan 31 at 17:18
mcaiojethewomcaiojethewo
477
$endgroup$
add a comment |
$begingroup$
I'm trying to teach myself separation of variables and have been following some notes for the wave equation, but there's one part which really confuses me and I'm not exactly sure how it makes the step.
For the wave equation
$$u_{tt} - c^2 u_{xx} = 0$$
with length $l$ and fixed ends, $u(0,t)=u(l,t)=0$ we seek a solution in the form $$u(x,t)=v(x)q(t)$$
and substituting this into the equation gives
$$frac{1}{q(t)}frac{d^2 q}{dt^2}=c^2 frac{1}{v(x)}frac{d^2 v}{dx^2}=-omega ^2$$
and I understand all of this so far. But when solving the equation
$$q''+omega ^2 q = 0$$
the general solution is
$$q(t)=A cos(omega t +alpha)$$
where $A$ and $alpha$ are constants - which is the part which I'm not exactly sure on how it gets to.
Could someone explain this step to me? Thanks!
pde wave-equation
edited Jan 31 at 17:42
Adrian Keister
5,26971933
asked Jan 31 at 17:18
mcaiojethewomcaiojethewo
477
$endgroup$
$begingroup$
There are several different ways to write the solution of that second-order ODE: $Asin(omega t)+Bcos(omega t),$ or $Acos(omega t+alpha),$ or $A e^{iomega t}+Be^{-iomega t}.$ The angle offset ends up being mathematically equivalent to another basis function (need two for the general solution to a linear second-order ODE). Does that help?
$endgroup$
– Adrian Keister
Jan 31 at 17:28
$begingroup$
Ooh okay, it didn't occur to me that it was just a different way of writing the other solutions. Is there a reason why it's advantageous to write it like that in this case?
$endgroup$
– mcaiojethewo
Jan 31 at 17:36
$begingroup$
Sometimes! It depends on where you want to go with it. In this case, the nice thing is that it's only one function to write.
$endgroup$
– Adrian Keister
Jan 31 at 17:37
$begingroup$
Ah okay, thank you for your help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:40
add a comment |
$begingroup$
I'm trying to teach myself separation of variables and have been following some notes for the wave equation, but there's one part which really confuses me and I'm not exactly sure how it makes the step.
For the wave equation
$$u_{tt} - c^2 u_{xx} = 0$$
with length $l$ and fixed ends, $u(0,t)=u(l,t)=0$ we seek a solution in the form $$u(x,t)=v(x)q(t)$$
and substituting this into the equation gives
$$frac{1}{q(t)}frac{d^2 q}{dt^2}=c^2 frac{1}{v(x)}frac{d^2 v}{dx^2}=-omega ^2$$
and I understand all of this so far. But when solving the equation
$$q''+omega ^2 q = 0$$
the general solution is
$$q(t)=A cos(omega t +alpha)$$
where $A$ and $alpha$ are constants - which is the part which I'm not exactly sure on how it gets to.
Could someone explain this step to me? Thanks!
pde wave-equation
edited Jan 31 at 17:42
Adrian Keister
5,26971933
asked Jan 31 at 17:18
mcaiojethewomcaiojethewo
477
$endgroup$
I'm trying to teach myself separation of variables and have been following some notes for the wave equation, but there's one part which really confuses me and I'm not exactly sure how it makes the step.
For the wave equation
$$u_{tt} - c^2 u_{xx} = 0$$
with length $l$ and fixed ends, $u(0,t)=u(l,t)=0$ we seek a solution in the form $$u(x,t)=v(x)q(t)$$
and substituting this into the equation gives
$$frac{1}{q(t)}frac{d^2 q}{dt^2}=c^2 frac{1}{v(x)}frac{d^2 v}{dx^2}=-omega ^2$$
and I understand all of this so far. But when solving the equation
$$q''+omega ^2 q = 0$$
the general solution is
$$q(t)=A cos(omega t +alpha)$$
where $A$ and $alpha$ are constants - which is the part which I'm not exactly sure on how it gets to.
Could someone explain this step to me? Thanks!
pde wave-equation
pde wave-equation
edited Jan 31 at 17:42
Adrian Keister
5,26971933
asked Jan 31 at 17:18
mcaiojethewomcaiojethewo
477
edited Jan 31 at 17:42
Adrian Keister
5,26971933
asked Jan 31 at 17:18
mcaiojethewomcaiojethewo
477
edited Jan 31 at 17:42
Adrian Keister
5,26971933
edited Jan 31 at 17:42
Adrian Keister
5,26971933
edited Jan 31 at 17:42
Adrian Keister
5,26971933
5,26971933
asked Jan 31 at 17:18
mcaiojethewomcaiojethewo
477
asked Jan 31 at 17:18
mcaiojethewomcaiojethewo
477
asked Jan 31 at 17:18
mcaiojethewomcaiojethewo
477
477
$begingroup$
There are several different ways to write the solution of that second-order ODE: $Asin(omega t)+Bcos(omega t),$ or $Acos(omega t+alpha),$ or $A e^{iomega t}+Be^{-iomega t}.$ The angle offset ends up being mathematically equivalent to another basis function (need two for the general solution to a linear second-order ODE). Does that help?
$endgroup$
– Adrian Keister
Jan 31 at 17:28
$begingroup$
Ooh okay, it didn't occur to me that it was just a different way of writing the other solutions. Is there a reason why it's advantageous to write it like that in this case?
$endgroup$
– mcaiojethewo
Jan 31 at 17:36
$begingroup$
Sometimes! It depends on where you want to go with it. In this case, the nice thing is that it's only one function to write.
$endgroup$
– Adrian Keister
Jan 31 at 17:37
$begingroup$
Ah okay, thank you for your help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:40
add a comment |
$begingroup$
There are several different ways to write the solution of that second-order ODE: $Asin(omega t)+Bcos(omega t),$ or $Acos(omega t+alpha),$ or $A e^{iomega t}+Be^{-iomega t}.$ The angle offset ends up being mathematically equivalent to another basis function (need two for the general solution to a linear second-order ODE). Does that help?
$endgroup$
– Adrian Keister
Jan 31 at 17:28
$begingroup$
Ooh okay, it didn't occur to me that it was just a different way of writing the other solutions. Is there a reason why it's advantageous to write it like that in this case?
$endgroup$
– mcaiojethewo
Jan 31 at 17:36
$begingroup$
Sometimes! It depends on where you want to go with it. In this case, the nice thing is that it's only one function to write.
$endgroup$
– Adrian Keister
Jan 31 at 17:37
$begingroup$
Ah okay, thank you for your help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:40
$begingroup$
There are several different ways to write the solution of that second-order ODE: $Asin(omega t)+Bcos(omega t),$ or $Acos(omega t+alpha),$ or $A e^{iomega t}+Be^{-iomega t}.$ The angle offset ends up being mathematically equivalent to another basis function (need two for the general solution to a linear second-order ODE). Does that help?
$endgroup$
– Adrian Keister
Jan 31 at 17:28
$begingroup$
There are several different ways to write the solution of that second-order ODE: $Asin(omega t)+Bcos(omega t),$ or $Acos(omega t+alpha),$ or $A e^{iomega t}+Be^{-iomega t}.$ The angle offset ends up being mathematically equivalent to another basis function (need two for the general solution to a linear second-order ODE). Does that help?
$endgroup$
– Adrian Keister
Jan 31 at 17:28
$begingroup$
Ooh okay, it didn't occur to me that it was just a different way of writing the other solutions. Is there a reason why it's advantageous to write it like that in this case?
$endgroup$
– mcaiojethewo
Jan 31 at 17:36
$begingroup$
Ooh okay, it didn't occur to me that it was just a different way of writing the other solutions. Is there a reason why it's advantageous to write it like that in this case?
$endgroup$
– mcaiojethewo
Jan 31 at 17:36
$begingroup$
Sometimes! It depends on where you want to go with it. In this case, the nice thing is that it's only one function to write.
$endgroup$
– Adrian Keister
Jan 31 at 17:37
$begingroup$
Sometimes! It depends on where you want to go with it. In this case, the nice thing is that it's only one function to write.
$endgroup$
– Adrian Keister
Jan 31 at 17:37
$begingroup$
Ah okay, thank you for your help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:40
$begingroup$
Ah okay, thank you for your help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:40
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Generally, it is not a good idea to do "Partial Differential Equations" until you have mastered "Ordinary Differential Equations"!
In "Ordinary Differential Equations" you learned that the "characteristic equation" of the differential equation $q''+ omega^2q= 0$ is $m^2+ omega^2= 0$ which, assuming $omega$ is real, has roots $m= omega i$. The general solution to that differential equation is $q= Ae^{omega ti}+ Be^{-omega ti}= A(cos(omega t)+ sin(omega t))+ B(cos(omega t)- sin(omega t))= (A+ B)cos(omega t)+ (A- B)sin(omega t)$.
In order to get the form you have, you need the trig identity $cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)$. So $A cos(omega t+ alpha)= Acos(alpha) cos(omega t)- Asin(alpha) sin(omega t)$.
edited Jan 31 at 17:38
Adrian Keister
5,26971933
answered Jan 31 at 17:35
user247327user247327
11.5k1516
$endgroup$
$begingroup$
It was just a case of me being stupid and not realising that it was just written in a different form than I was used to - thanks for the help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:39
$begingroup$
@mcaiojethewo: This is not stupidity! Stupidity, in this case, would look like not caring and not asking.
$endgroup$
– Adrian Keister
Jan 31 at 17:43
add a comment |
$begingroup$
So lets try to work through this:
Intuitively, you can see that cosine is the correct solution since by differentiating twice we get back to the cosine but only its negative. However, we can do this more formally(this is my favourite method of doing it):
Suppose we can get a solution of the sort $u(x)=sum^{infty}_{i=1}a_ix^i$.Then we can write:
$u'(x)=sum^{infty}_{i=1}ia_ix^{i-1}=sum^{infty}_{i=0}(i+1)a_{i+1}x^{i}$, and similarly :
$u''(x)=sum^{infty}_{i=0}(i+1)(i)a_{i+1}x^{i-1}=sum^{infty}_{i=0}(i+1)(i+2)a_{i+2}x^{i}$.
So we now have the following:
$sum^{infty}_{i=0}(i+1)(i+2)a_{i+2}x^{i}+w^2a_ix^i=0$.
This will give you that $a_{i+2}=-(w^2)a_i/(i+1)(i+2)$.
If you solve this linear recurence relation,with the initial conditions, you will get the power series for cosine
answered Jan 31 at 17:39
Marat AlievMarat Aliev
1312
$endgroup$
add a comment |
$begingroup$
I would get rid of $alpha$ and just look at your boundary conditions to solve your constants. The first thing to note, is that you should begin in fact with:
begin{equation}
sum_{n=1}^{infty} A_ncos(omega_nt) + B_nsin(omega_nt)
end{equation}
From there you can use boundary conditions, signs and fourier series knowledge to end up with only cos and fill in the coefficients you're missing.
edited Jan 31 at 17:41
Adrian Keister
5,26971933
answered Jan 31 at 17:39
EMPEMP
83
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Generally, it is not a good idea to do "Partial Differential Equations" until you have mastered "Ordinary Differential Equations"!
In "Ordinary Differential Equations" you learned that the "characteristic equation" of the differential equation $q''+ omega^2q= 0$ is $m^2+ omega^2= 0$ which, assuming $omega$ is real, has roots $m= omega i$. The general solution to that differential equation is $q= Ae^{omega ti}+ Be^{-omega ti}= A(cos(omega t)+ sin(omega t))+ B(cos(omega t)- sin(omega t))= (A+ B)cos(omega t)+ (A- B)sin(omega t)$.
In order to get the form you have, you need the trig identity $cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)$. So $A cos(omega t+ alpha)= Acos(alpha) cos(omega t)- Asin(alpha) sin(omega t)$.
edited Jan 31 at 17:38
Adrian Keister
5,26971933
answered Jan 31 at 17:35
user247327user247327
11.5k1516
$endgroup$
$begingroup$
It was just a case of me being stupid and not realising that it was just written in a different form than I was used to - thanks for the help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:39
$begingroup$
@mcaiojethewo: This is not stupidity! Stupidity, in this case, would look like not caring and not asking.
$endgroup$
– Adrian Keister
Jan 31 at 17:43
add a comment |
$begingroup$
Generally, it is not a good idea to do "Partial Differential Equations" until you have mastered "Ordinary Differential Equations"!
In "Ordinary Differential Equations" you learned that the "characteristic equation" of the differential equation $q''+ omega^2q= 0$ is $m^2+ omega^2= 0$ which, assuming $omega$ is real, has roots $m= omega i$. The general solution to that differential equation is $q= Ae^{omega ti}+ Be^{-omega ti}= A(cos(omega t)+ sin(omega t))+ B(cos(omega t)- sin(omega t))= (A+ B)cos(omega t)+ (A- B)sin(omega t)$.
In order to get the form you have, you need the trig identity $cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)$. So $A cos(omega t+ alpha)= Acos(alpha) cos(omega t)- Asin(alpha) sin(omega t)$.
edited Jan 31 at 17:38
Adrian Keister
5,26971933
answered Jan 31 at 17:35
user247327user247327
11.5k1516
$endgroup$
$begingroup$
It was just a case of me being stupid and not realising that it was just written in a different form than I was used to - thanks for the help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:39
$begingroup$
@mcaiojethewo: This is not stupidity! Stupidity, in this case, would look like not caring and not asking.
$endgroup$
– Adrian Keister
Jan 31 at 17:43
add a comment |
$begingroup$
Generally, it is not a good idea to do "Partial Differential Equations" until you have mastered "Ordinary Differential Equations"!
In "Ordinary Differential Equations" you learned that the "characteristic equation" of the differential equation $q''+ omega^2q= 0$ is $m^2+ omega^2= 0$ which, assuming $omega$ is real, has roots $m= omega i$. The general solution to that differential equation is $q= Ae^{omega ti}+ Be^{-omega ti}= A(cos(omega t)+ sin(omega t))+ B(cos(omega t)- sin(omega t))= (A+ B)cos(omega t)+ (A- B)sin(omega t)$.
In order to get the form you have, you need the trig identity $cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)$. So $A cos(omega t+ alpha)= Acos(alpha) cos(omega t)- Asin(alpha) sin(omega t)$.
edited Jan 31 at 17:38
Adrian Keister
5,26971933
answered Jan 31 at 17:35
user247327user247327
11.5k1516
$endgroup$
Generally, it is not a good idea to do "Partial Differential Equations" until you have mastered "Ordinary Differential Equations"!
In "Ordinary Differential Equations" you learned that the "characteristic equation" of the differential equation $q''+ omega^2q= 0$ is $m^2+ omega^2= 0$ which, assuming $omega$ is real, has roots $m= omega i$. The general solution to that differential equation is $q= Ae^{omega ti}+ Be^{-omega ti}= A(cos(omega t)+ sin(omega t))+ B(cos(omega t)- sin(omega t))= (A+ B)cos(omega t)+ (A- B)sin(omega t)$.
In order to get the form you have, you need the trig identity $cos(A+ B)= cos(A)cos(B)- sin(A)sin(B)$. So $A cos(omega t+ alpha)= Acos(alpha) cos(omega t)- Asin(alpha) sin(omega t)$.
edited Jan 31 at 17:38
Adrian Keister
5,26971933
answered Jan 31 at 17:35
user247327user247327
11.5k1516
edited Jan 31 at 17:38
Adrian Keister
5,26971933
edited Jan 31 at 17:38
Adrian Keister
5,26971933
edited Jan 31 at 17:38
Adrian Keister
5,26971933
5,26971933
answered Jan 31 at 17:35
user247327user247327
11.5k1516
answered Jan 31 at 17:35
user247327user247327
11.5k1516
answered Jan 31 at 17:35
user247327user247327
11.5k1516
11.5k1516
$begingroup$
It was just a case of me being stupid and not realising that it was just written in a different form than I was used to - thanks for the help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:39
$begingroup$
@mcaiojethewo: This is not stupidity! Stupidity, in this case, would look like not caring and not asking.
$endgroup$
– Adrian Keister
Jan 31 at 17:43
add a comment |
$begingroup$
It was just a case of me being stupid and not realising that it was just written in a different form than I was used to - thanks for the help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:39
$begingroup$
@mcaiojethewo: This is not stupidity! Stupidity, in this case, would look like not caring and not asking.
$endgroup$
– Adrian Keister
Jan 31 at 17:43
$begingroup$
It was just a case of me being stupid and not realising that it was just written in a different form than I was used to - thanks for the help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:39
$begingroup$
It was just a case of me being stupid and not realising that it was just written in a different form than I was used to - thanks for the help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:39
$begingroup$
@mcaiojethewo: This is not stupidity! Stupidity, in this case, would look like not caring and not asking.
$endgroup$
– Adrian Keister
Jan 31 at 17:43
$begingroup$
@mcaiojethewo: This is not stupidity! Stupidity, in this case, would look like not caring and not asking.
$endgroup$
– Adrian Keister
Jan 31 at 17:43
add a comment |
$begingroup$
So lets try to work through this:
Intuitively, you can see that cosine is the correct solution since by differentiating twice we get back to the cosine but only its negative. However, we can do this more formally(this is my favourite method of doing it):
Suppose we can get a solution of the sort $u(x)=sum^{infty}_{i=1}a_ix^i$.Then we can write:
$u'(x)=sum^{infty}_{i=1}ia_ix^{i-1}=sum^{infty}_{i=0}(i+1)a_{i+1}x^{i}$, and similarly :
$u''(x)=sum^{infty}_{i=0}(i+1)(i)a_{i+1}x^{i-1}=sum^{infty}_{i=0}(i+1)(i+2)a_{i+2}x^{i}$.
So we now have the following:
$sum^{infty}_{i=0}(i+1)(i+2)a_{i+2}x^{i}+w^2a_ix^i=0$.
This will give you that $a_{i+2}=-(w^2)a_i/(i+1)(i+2)$.
If you solve this linear recurence relation,with the initial conditions, you will get the power series for cosine
answered Jan 31 at 17:39
Marat AlievMarat Aliev
1312
$endgroup$
add a comment |
$begingroup$
So lets try to work through this:
Intuitively, you can see that cosine is the correct solution since by differentiating twice we get back to the cosine but only its negative. However, we can do this more formally(this is my favourite method of doing it):
Suppose we can get a solution of the sort $u(x)=sum^{infty}_{i=1}a_ix^i$.Then we can write:
$u'(x)=sum^{infty}_{i=1}ia_ix^{i-1}=sum^{infty}_{i=0}(i+1)a_{i+1}x^{i}$, and similarly :
$u''(x)=sum^{infty}_{i=0}(i+1)(i)a_{i+1}x^{i-1}=sum^{infty}_{i=0}(i+1)(i+2)a_{i+2}x^{i}$.
So we now have the following:
$sum^{infty}_{i=0}(i+1)(i+2)a_{i+2}x^{i}+w^2a_ix^i=0$.
This will give you that $a_{i+2}=-(w^2)a_i/(i+1)(i+2)$.
If you solve this linear recurence relation,with the initial conditions, you will get the power series for cosine
answered Jan 31 at 17:39
Marat AlievMarat Aliev
1312
$endgroup$
add a comment |
$begingroup$
So lets try to work through this:
Intuitively, you can see that cosine is the correct solution since by differentiating twice we get back to the cosine but only its negative. However, we can do this more formally(this is my favourite method of doing it):
Suppose we can get a solution of the sort $u(x)=sum^{infty}_{i=1}a_ix^i$.Then we can write:
$u'(x)=sum^{infty}_{i=1}ia_ix^{i-1}=sum^{infty}_{i=0}(i+1)a_{i+1}x^{i}$, and similarly :
$u''(x)=sum^{infty}_{i=0}(i+1)(i)a_{i+1}x^{i-1}=sum^{infty}_{i=0}(i+1)(i+2)a_{i+2}x^{i}$.
So we now have the following:
$sum^{infty}_{i=0}(i+1)(i+2)a_{i+2}x^{i}+w^2a_ix^i=0$.
This will give you that $a_{i+2}=-(w^2)a_i/(i+1)(i+2)$.
If you solve this linear recurence relation,with the initial conditions, you will get the power series for cosine
answered Jan 31 at 17:39
Marat AlievMarat Aliev
1312
$endgroup$
So lets try to work through this:
Intuitively, you can see that cosine is the correct solution since by differentiating twice we get back to the cosine but only its negative. However, we can do this more formally(this is my favourite method of doing it):
Suppose we can get a solution of the sort $u(x)=sum^{infty}_{i=1}a_ix^i$.Then we can write:
$u'(x)=sum^{infty}_{i=1}ia_ix^{i-1}=sum^{infty}_{i=0}(i+1)a_{i+1}x^{i}$, and similarly :
$u''(x)=sum^{infty}_{i=0}(i+1)(i)a_{i+1}x^{i-1}=sum^{infty}_{i=0}(i+1)(i+2)a_{i+2}x^{i}$.
So we now have the following:
$sum^{infty}_{i=0}(i+1)(i+2)a_{i+2}x^{i}+w^2a_ix^i=0$.
This will give you that $a_{i+2}=-(w^2)a_i/(i+1)(i+2)$.
If you solve this linear recurence relation,with the initial conditions, you will get the power series for cosine
answered Jan 31 at 17:39
Marat AlievMarat Aliev
1312
answered Jan 31 at 17:39
Marat AlievMarat Aliev
1312
answered Jan 31 at 17:39
Marat AlievMarat Aliev
1312
answered Jan 31 at 17:39
Marat AlievMarat Aliev
1312
1312
add a comment |
add a comment |
$begingroup$
I would get rid of $alpha$ and just look at your boundary conditions to solve your constants. The first thing to note, is that you should begin in fact with:
begin{equation}
sum_{n=1}^{infty} A_ncos(omega_nt) + B_nsin(omega_nt)
end{equation}
From there you can use boundary conditions, signs and fourier series knowledge to end up with only cos and fill in the coefficients you're missing.
edited Jan 31 at 17:41
Adrian Keister
5,26971933
answered Jan 31 at 17:39
EMPEMP
83
$endgroup$
add a comment |
$begingroup$
I would get rid of $alpha$ and just look at your boundary conditions to solve your constants. The first thing to note, is that you should begin in fact with:
begin{equation}
sum_{n=1}^{infty} A_ncos(omega_nt) + B_nsin(omega_nt)
end{equation}
From there you can use boundary conditions, signs and fourier series knowledge to end up with only cos and fill in the coefficients you're missing.
edited Jan 31 at 17:41
Adrian Keister
5,26971933
answered Jan 31 at 17:39
EMPEMP
83
$endgroup$
add a comment |
$begingroup$
I would get rid of $alpha$ and just look at your boundary conditions to solve your constants. The first thing to note, is that you should begin in fact with:
begin{equation}
sum_{n=1}^{infty} A_ncos(omega_nt) + B_nsin(omega_nt)
end{equation}
From there you can use boundary conditions, signs and fourier series knowledge to end up with only cos and fill in the coefficients you're missing.
edited Jan 31 at 17:41
Adrian Keister
5,26971933
answered Jan 31 at 17:39
EMPEMP
83
$endgroup$
I would get rid of $alpha$ and just look at your boundary conditions to solve your constants. The first thing to note, is that you should begin in fact with:
begin{equation}
sum_{n=1}^{infty} A_ncos(omega_nt) + B_nsin(omega_nt)
end{equation}
From there you can use boundary conditions, signs and fourier series knowledge to end up with only cos and fill in the coefficients you're missing.
edited Jan 31 at 17:41
Adrian Keister
5,26971933
answered Jan 31 at 17:39
EMPEMP
83
edited Jan 31 at 17:41
Adrian Keister
5,26971933
edited Jan 31 at 17:41
Adrian Keister
5,26971933
edited Jan 31 at 17:41
Adrian Keister
5,26971933
5,26971933
answered Jan 31 at 17:39
EMPEMP
83
answered Jan 31 at 17:39
EMPEMP
83
answered Jan 31 at 17:39
EMPEMP
83
83
add a comment |
add a comment |
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$begingroup$
There are several different ways to write the solution of that second-order ODE: $Asin(omega t)+Bcos(omega t),$ or $Acos(omega t+alpha),$ or $A e^{iomega t}+Be^{-iomega t}.$ The angle offset ends up being mathematically equivalent to another basis function (need two for the general solution to a linear second-order ODE). Does that help?
$endgroup$
– Adrian Keister
Jan 31 at 17:28
$begingroup$
Ooh okay, it didn't occur to me that it was just a different way of writing the other solutions. Is there a reason why it's advantageous to write it like that in this case?
$endgroup$
– mcaiojethewo
Jan 31 at 17:36
$begingroup$
Sometimes! It depends on where you want to go with it. In this case, the nice thing is that it's only one function to write.
$endgroup$
– Adrian Keister
Jan 31 at 17:37
$begingroup$
Ah okay, thank you for your help!
$endgroup$
– mcaiojethewo
Jan 31 at 17:40