Linear Equation with Constant Coefficients Problem

[JM00404]

Find all solutions \phi of y''+y=0 satisfying:
1) \phi(0)=1, \phi(\pi/2)=2
2) \phi(0)=0, \phi(\pi)=0
3) \phi(0)=0, \phi'(\pi/2)=0
4) \phi(0)=0, \phi(\pi/2)=0

I cannot seem to solve parts 2-4 in a way that would result in the coefficients being any constant other than zero. Whenever I solve for one of the constants in the equation \phi(0)=0 , I find that C_1=-C_2 ; which isn't alarming yet since the magnitude of either constant is not yet known. When I try to solve the second property in each of the subsequent equations, I find that C_2=0 , for example, which implies that C_1=0 since C_1=-C_2 . Below, I have included a portion of the work that I completed to get to the current situation that I am in. Any assistance offered would be much appreciated. Thank you for your time.

y''+y=0 .
Characteristic Polynomial: p(r)=r^2+1=(r-i)(r+i)
\implies Roots =\pm i .
\therefore \phi(x)=C_1e^ {ix} +C_2e^ {-ix} where C_1 & C_2 are constants.
\phi(0)=C_1e^0+C_2e^0=C_1+C_2=1 \implies C_1=1-C_2 .
\phi(\pi/2)=C_1e^ {\pi i/2} +C_2e^{ -\pi i/2} =(1-C_2)e^ {\pi i/2} +C_2e^ {-\pi i/2} =\cdots=(1-2C_2)i=2 \implies C_2=(\frac 1 2 +i) \implies C_1=(\frac 1 2 -i ).
\therefore \phi(x)=(\frac 1 2 -i)e^ {ix} +(\frac 1 2 +i)e^ {-ix} =\cdots=cos(x)+2sin(x) .

 

[George Jones]

The first condition of 2) gives, as you say, C_1 = -C_2. Using this and the second condition of 2), I get that C_2 can be anything.

Regards,
George

 

[hunchback6116]

b) y=(c1)sinx
c) y=(c1)sinx
d) y=o

(c1)= any constant

For b and c, I get (c2)=0, so (c1) can be any constant and there is no cosx term
For d, I get both (c1) and (c2) are 0, so the only soln. is y=0

 

[George Jones]

b) y=(c1)sinx
c) y=(c1)sinx
d) y=o

(c1)= any constant

For b and c, I get (c2)=0, so (c1) can be any constant and there is no cosx term
For d, I get both (c1) and (c2) are 0, so the only soln. is y=0

Careful - I think your C's are different JM00404's C's. This doesn't affect the final answers, but it does affect the values of the C's.

JM00404 has chose C's such that

\phi(x) = C_1 e^{ix} + C_2 e^{-ix}.

I think your C's are such that

\phi(x) = C_1 sinx + C_2 cosx.

Regards,
George

 

[hunchback6116]

You're right

You're right. The general solution to the equation that I am using is

y(x) = (c1)sinx + (c2)cosx

 

[shaner-baner]

Lets look @ (2). we have a general solution y=(c1)sinx+(c2)cosx, the first restriction y(0)=0, inplies that c2=0. The second restriction y(Pi)=0 does nothing because for any value of c1, sinx will be zero at Pi. so the solution is a bunch of sine curves with any amplitude. This is a good example of a second order d.e. that has 2 boundary conditions that don't determine a unique solution.