Higher-Order Linear DE Problem!

Tom999

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Feb 27, 2019
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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.
 
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. :)
 
Yes, I'm familiar with the method. I'm a bit frazzled by the execution though.
 
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?
 
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 v1(x)v_1(x) and v2(x)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 v1v_1' and v2v_2'?
 
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The first equation implies:

[MATH]v_1'=-\tan(wx)v_2'[/MATH]
And then substituting for v1v_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)\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 ww, 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 v1(x)v_1(x) by integrating?
 
Yes, I can do that from there. Thank you!

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)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.
 
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