JM00404
- 7
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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) .
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) .