Linear Equation with Constant Coefficients Problem

AI Thread Summary
The discussion revolves around solving the linear differential equation y'' + y = 0 with various boundary conditions. For the first condition, the solution is found to be φ(x) = cos(x) + 2sin(x) with specific constants derived from the initial conditions. However, parts 2 to 4 lead to the conclusion that the constants C1 and C2 must be zero, resulting in the trivial solution y = 0 for some cases. The general solution for conditions 2 and 3 indicates that C2 is zero, allowing C1 to be any constant, which means multiple sine curves are valid solutions. Overall, the boundary conditions do not yield unique solutions for all cases discussed.
JM00404
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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) .
 
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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
 
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
 
hunchback6116 said:
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
 
You're right

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

y(x) = (c1)sinx + (c2)cosx
 
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.
 
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