Find all solutions [tex] \phi [/tex] of [tex] y''+y=0 [/tex] satisfying:(adsbygoogle = window.adsbygoogle || []).push({});

1) [tex] \phi(0)=1, \phi(\pi/2)=2 [/tex]

2) [tex] \phi(0)=0, \phi(\pi)=0 [/tex]

3) [tex] \phi(0)=0, \phi'(\pi/2)=0 [/tex]

4) [tex] \phi(0)=0, \phi(\pi/2)=0 [/tex]

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 [tex] \phi(0)=0 [/tex] , I find that [tex] C_1=-C_2 [/tex] ; 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 [tex] C_2=0 [/tex] , for example, which implies that [tex] C_1=0 [/tex] since [tex] C_1=-C_2 [/tex] . 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.

[tex] y''+y=0 [/tex] .

Characteristic Polynomial: [tex] p(r)=r^2+1=(r-i)(r+i) [/tex]

[tex] \implies [/tex] Roots [tex] =\pm i [/tex] .

[tex] \therefore \phi(x)=C_1e^ {ix} +C_2e^ {-ix} [/tex] where [tex] C_1 [/tex] & [tex] C_2 [/tex] are constants.

[tex] \phi(0)=C_1e^0+C_2e^0=C_1+C_2=1 \implies C_1=1-C_2 [/tex] .

[tex] \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 ) [/tex].

[tex] \therefore \phi(x)=(\frac 1 2 -i)e^ {ix} +(\frac 1 2 +i)e^ {-ix} =\cdots=cos(x)+2sin(x) [/tex] .

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# Homework Help: Linear Equation with Constant Coefficients Problem

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