# Linear Equation with Constant Coefficients Problem

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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George Jones
Staff Emeritus
Gold Member
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

George Jones
Staff Emeritus
Gold Member
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.