Question

Solve y''+y=cot(x).

I know I asked a similar, I had the signs incorrect, my bad. Ok from here I have the characteristic equation is r^2+1=0 which gives r=i,-i. this give y1=cosx and y2=sinx.....right? Where do I go from here?

Answer 1

Like last time, I'm pretty sure you'll need to use variation of parameters. I haven't come across a solution to an equation like this that did not involve VoP, other than (maybe) power series.

Answer 2

Apparently we DID learn it but he never ever gave it a name in class. I had to find it in the book and it's basically like what he did in examples. It was right after undetermined coefficients. I'm sorry I didn't realize that yesterday!

Answer 3

That's okay! First thing you do is check the fundamental solutions \(\sin x\) and \(\cos x\) are linearly independent. Do you know how to do that with the Wronskian?

Answer 4

[cosx sinx
-sinx cosx] and then take the determinant?

Answer 5

and the determinant=1 which isn't 0 so they are linearly independent....?

Answer 6

Right.

Answer 7

Now you use the formulas for the new-parameter solution,
\[u_1=-\int\frac{y_2g(t)}{W(y_1,y_2)}~dt~~~~\text{and}~~~~u_2=\int\frac{y_2g(t)}{W(y_1,y_2)}~dt\]
where \(y_1=\cos x\), \(y_2=\sin x\), \(W(y_1,y_2)\) is the Wronskian, and \(g(t)=\cot x\).

The final solution will be
\[y=u_1y_1+u_2y_2\]

Answer 8

oh man....the answer's gonna be ugly isn't it?

Answer 9

Surprisingly, no! Well, maybe a little, but one of those integrals will simplify to something really nice.

Answer 10

should that be y1 in the formula for u1?

Answer 11

Nope, \(u_1\) is right, \(u_2\) is wrong:
\[u_1=-\int\frac{y_2g(t)}{W(y_1,y_2)}~dt~~~~\text{and}~~~~u_2=\int\frac{\color{red}{y_1}g(t)}{W(y_1,y_2)}~dt\]
Thanks for catching that.

Answer 12

so.....sorry I'm just making sure I have all steps correct.....u1 should become -sin(x)...and u2= cos(x) +log(sin(x/2))-log(cos(x/2))

Answer 13

and the final, general solution will be c1cosx +c2sinx + (u1y1 +u2y2)?

Answer 14

\(u_1\) is correct. I don't have any paper at hand so I'll check your work with the second integral:
\[\begin{align*}u_2&=\int\cos x\cot x~dx\\&=\cot x\sin x+\int\csc^2x\sin x~dx&\text{integrating by parts}\\
&=\cos x+\int\csc x~dx\\
&=\cos x-\ln\left|\csc x+\cot x\right|
\end{align*}\]
With some identities, you'd get
\[u_2=\cos x+\ln\left|\sin\frac{x}{2}\right|-\ln\left|\cos\frac{x}{2}\right|\]
So yes, you're right.

And yes, the general solution would be as you wrote it.

Answer 15

AWESOME!!!! thank you so so so much!

Answer 16

You're welcome!