Higher-Order Linear DE Problem!

[Tom999]

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]

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]

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

 

[MarkFL]

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]

y = (C1)coswx + (C2)sinwx

 

[MarkFL]

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 $v_2'$?

 

[Tom999]

I'm not sure how to proceed from there.

 

[MarkFL]

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?

 

[Tom999]

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

 

[MarkFL]

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 $g(x)=\tan(qx)$...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.