Higher-Order Linear DE Problem!
- Thread starter Tom999
- Start date
- Joined
- Nov 24, 2012
- Messages
- 3,021
Yes, the method of undetermined coefficients isn't going to work here. Are you familiar with the method of variation of parameters? I suspect that's just what we need here, but before we proceed, I want to be sure you have heard of this method. 
- Joined
- Nov 24, 2012
- Messages
- 3,021
Okay, good. First, we need the homogeneous solution...what do you have for that?
- Joined
- Nov 24, 2012
- Messages
- 3,021
Yes, we have:
[MATH]y_h(x)=c_1\cos(wx)+c_2\sin(wx)[/MATH]
And so, we seek to find a particular solution of the form:
[MATH]y_p(x)=v_1(x)\cos(wx)+v_2(x)\sin(wx)[/MATH]
So, we need to determine \(v_1(x)\) and \(v_2(x)\), by solving the following system and integrating:
[MATH]\cos(wx)v_1'+\sin(wx)v_2'=0[/MATH]
[MATH]-w\sin(wx)v_1'+w\cos(wx)v_2'=\tan(wx)[/MATH]
Can you solve the above system for \(v_1'\) and \(v_2'\)?
[MATH]y_h(x)=c_1\cos(wx)+c_2\sin(wx)[/MATH]
And so, we seek to find a particular solution of the form:
[MATH]y_p(x)=v_1(x)\cos(wx)+v_2(x)\sin(wx)[/MATH]
So, we need to determine \(v_1(x)\) and \(v_2(x)\), by solving the following system and integrating:
[MATH]\cos(wx)v_1'+\sin(wx)v_2'=0[/MATH]
[MATH]-w\sin(wx)v_1'+w\cos(wx)v_2'=\tan(wx)[/MATH]
Can you solve the above system for \(v_1'\) and \(v_2'\)?
Last edited:
- Joined
- Nov 24, 2012
- Messages
- 3,021
The first equation implies:
[MATH]v_1'=-\tan(wx)v_2'[/MATH]
And then substituting for \(v_1'\) into the second equation we get:
[MATH]w\sin(wx)\tan(wx)v_2'+w\cos(wx)v_2'=\tan(wx)[/MATH]
Multiply through by \(\cos(wx)\) to get:
[MATH]w\sin^2(wx)v_2'+w\cos^2(wx)v_2'=\sin(wx)[/MATH]
Apply a Pythagorean identity to the LHS, divided through by \(w\), and we have:
[MATH]v_2'=\frac{1}{w}\sin(wx)\implies v_2(x)=-\frac{1}{w^2}\cos(wx)[/MATH]
And so:
[MATH]v_1'=-\frac{1}{w}\sin(wx)\tan(wx)[/MATH]
Can you now find \(v_1(x)\) by integrating?
[MATH]v_1'=-\tan(wx)v_2'[/MATH]
And then substituting for \(v_1'\) into the second equation we get:
[MATH]w\sin(wx)\tan(wx)v_2'+w\cos(wx)v_2'=\tan(wx)[/MATH]
Multiply through by \(\cos(wx)\) to get:
[MATH]w\sin^2(wx)v_2'+w\cos^2(wx)v_2'=\sin(wx)[/MATH]
Apply a Pythagorean identity to the LHS, divided through by \(w\), and we have:
[MATH]v_2'=\frac{1}{w}\sin(wx)\implies v_2(x)=-\frac{1}{w^2}\cos(wx)[/MATH]
And so:
[MATH]v_1'=-\frac{1}{w}\sin(wx)\tan(wx)[/MATH]
Can you now find \(v_1(x)\) by integrating?
- Joined
- Nov 24, 2012
- Messages
- 3,021
Just so you can check your result, I would write:
[MATH]v_1(x)=-\frac{1}{w^2}\int \sin(wx)\tan(wx)w\,dx=\frac{1}{w^2}\int \cos(wx)-\sec(wx)w\,dx=\frac{\sin(wx)-\ln(\sec(wx)+\tan(wx))}{w^2}[/MATH]
And so our particular solution is:
[MATH]y_p(x)=\left(\frac{\sin(wx)-\ln(\sec(wx)+\tan(wx))}{w^2}\right)\cos(wx)+\left(-\frac{1}{w^2}\cos(wx)\right)\sin(wx)=-\frac{1}{w^2}\cos(wx)\ln(\sec(wx)+\tan(wx))[/MATH]
And so, the general solution to the given ODE is:
[MATH]y(x)=y_h(x)+y_p(x)=c_1\cos(wx)+c_2\sin(wx)-\frac{1}{w^2}\cos(wx)\ln(\sec(wx)+\tan(wx))[/MATH]
And that would be our solution if I hadn't made a silly oversight...we have \(g(x)=\tan(qx)\)...and so we need to go all the way back to this system:
[MATH]\cos(wx)v_1'+\sin(wx)v_2'=0[/MATH]
[MATH]-w\sin(wx)v_1'+w\cos(wx)v_2'=\tan(qx)[/MATH]
As before, the first equation implies:
[MATH]v_1'=-\tan(wx)v_2'[/MATH]
And so the second equation becomes:
[MATH]-w\sin(wx)\left(-\tan(wx)v_2'\right)+w\cos(wx)v_2'=\tan(qx)[/MATH]
[MATH]w\sin^2(wx)v_2'+w\cos^2(wx)v_2'=\tan(qx)\cos(wx)[/MATH]
[MATH]v_2'=\frac{1}{w}\tan(qx)\cos(wx)[/MATH]
And I have no idea how to integrate that...W|A indicates that it involves the hypergeometric function.