Question

Solve the IVP: y''+y=0, y(pi/3)=0, y'(pi/3)=2. Find y(pi/6).

Answer 1

So this is linear and homogeneous.
Do you understand how to find the characteristic equation?

Answer 2

Oh also,
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Answer 3

Thanks

Answer 4

We assume our solution is of the form: \(\Large y=e^{rx}\)
Taking it's derivative twice gives us: \(\Large y''=r^2e^{rx}\)

Plugging these into our equation gives us:\[\Large r^2e^{rx}+e^{rx}=0\]Factoring gives us our characteristic equation:\[\Large e^{rx}(r^2+1)=0, \qquad\to\qquad r^2+1=0\]

Answer 5

Hmm so it looks like we end up with complex roots for the characteristic equation.
Do you remember what type of solutions we get when our roots are complex? :o

Answer 6

No

Answer 7

Roots of the form: \(\Large r\quad=\quad \alpha\pm\beta i\)
Give us a solution of the form: \(\Large y\quad=\quad c_1 e^{\alpha x}\cos\beta x+c_2e^{\alpha x}\sin\beta x\)Exponentials AND sines/cosines.

In our case, we have no real portion as part of our roots, so we won't get the exponentials.\[\Large r=\pm i\]Is what we got.

Answer 8

So our beta is 1.\[\Large y(x)\quad=\quad c_1\cos x+c_2 \sin x\]To solve for our unknown constants, we need to use the initial conditions that were given.

Answer 9

Oh I guess we need to find y'(x) at this point :o
Do you understand how to find the derivative here? :D

Answer 10

Too confusing? :c

Answer 11

DEs are a struggle for me

Answer 12

\[\Large y'(x)\quad=\quad ?\]So we ignore the constants.
Derivative of cosine?

Answer 13

-sin

Answer 14

\[\Large y'(x)\quad=\quad -c_1 \sin x+?\]Ok good and derivative of sine gives us cosine yes?\[\Large y'(x)\quad=\quad -c_1 \sin x+c_2 \cos x\]

Answer 15

So we've found the `general solution` and it's derivative:\[\Large y(x)\quad=\quad c_1\cos x+c_2 \sin x\]\[\Large y'(x)\quad=\quad -c_1 \sin x+c_2 \cos x\]

Answer 16

We want to use our initial data to come up with a system of equations:\[\Large y(\pi/3)\qquad\to\qquad c_1\cos(\pi/3)+c_2\sin(\pi/3)\quad=\quad 0\]\[\Large y'(\pi/3)\qquad\to\qquad -c_1\sin(\pi/3)+c_2\cos(\pi/3)=2\]