Higher-Order Linear DE Problem!

Tom999

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y'' + (w^2)y = tan(qx)

w and q are constants. I can find the solution to the Homo DE but am not sure how to approach the Inhomogeneous Part.
 

MarkFL

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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. :)
 

Tom999

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Yes, I'm familiar with the method. I'm a bit frazzled by the execution though.
 

MarkFL

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Yes, I'm familiar with the method. I'm a bit frazzled by the execution though.
Okay, good. First, we need the homogeneous solution...what do you have for that?
 

Tom999

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y = (C1)coswx + (C2)sinwx
 

MarkFL

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Yes, we have:

\(\displaystyle y_h(x)=c_1\cos(wx)+c_2\sin(wx)\)

And so, we seek to find a particular solution of the form:

\(\displaystyle y_p(x)=v_1(x)\cos(wx)+v_2(x)\sin(wx)\)

So, we need to determine \(v_1(x)\) and \(v_2(x)\), by solving the following system and integrating:

\(\displaystyle \cos(wx)v_1'+\sin(wx)v_2'=0\)

\(\displaystyle -w\sin(wx)v_1'+w\cos(wx)v_2'=\tan(wx)\)

Can you solve the above system for \(v_1'\) and \(v_2'\)?
 
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Tom999

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I'm not sure how to proceed from there.
 

MarkFL

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The first equation implies:

\(\displaystyle v_1'=-\tan(wx)v_2'\)

And then substituting for \(v_1'\) into the second equation we get:

\(\displaystyle w\sin(wx)\tan(wx)v_2'+w\cos(wx)v_2'=\tan(wx)\)

Multiply through by \(\cos(wx)\) to get:

\(\displaystyle w\sin^2(wx)v_2'+w\cos^2(wx)v_2'=\sin(wx)\)

Apply a Pythagorean identity to the LHS, divided through by \(w\), and we have:

\(\displaystyle v_2'=\frac{1}{w}\sin(wx)\implies v_2(x)=-\frac{1}{w^2}\cos(wx)\)

And so:

\(\displaystyle v_1'=-\frac{1}{w}\sin(wx)\tan(wx)\)

Can you now find \(v_1(x)\) by integrating?
 

Tom999

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Yes, I can do that from there. Thank you!
 

MarkFL

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Yes, I can do that from there. Thank you!
Just so you can check your result, I would write:

\(\displaystyle v_1(x)=-\frac{1}{w^2}\int \sin(wx)\tan(wx)w\,dx=\frac{1}{w^2}\int \cos(wx)-\sec(wx)w\,dx=\frac{\sin(wx)-\ln(\sec(wx)+\tan(wx))}{w^2}\)

And so our particular solution is:

\(\displaystyle y_p(x)=\left(\frac{\sin(wx)-\ln(\sec(wx)+\tan(wx))}{w^2}\right)\cos(wx)+\left(-\frac{1}{w^2}\cos(wx)\right)\sin(wx)=-\frac{1}{w^2}\cos(wx)\ln(\sec(wx)+\tan(wx))\)

And so, the general solution to the given ODE is:

\(\displaystyle y(x)=y_h(x)+y_p(x)=c_1\cos(wx)+c_2\sin(wx)-\frac{1}{w^2}\cos(wx)\ln(\sec(wx)+\tan(wx))\)

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:

\(\displaystyle \cos(wx)v_1'+\sin(wx)v_2'=0\)

\(\displaystyle -w\sin(wx)v_1'+w\cos(wx)v_2'=\tan(qx)\)

As before, the first equation implies:

\(\displaystyle v_1'=-\tan(wx)v_2'\)

And so the second equation becomes:

\(\displaystyle -w\sin(wx)\left(-\tan(wx)v_2'\right)+w\cos(wx)v_2'=\tan(qx)\)

\(\displaystyle w\sin^2(wx)v_2'+w\cos^2(wx)v_2'=\tan(qx)\cos(wx)\)

\(\displaystyle v_2'=\frac{1}{w}\tan(qx)\cos(wx)\)

And I have no idea how to integrate that...W|A indicates that it involves the hypergeometric function.
 
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