Which Differential Equation Has a Unique Solution on the Interval (0, pi)?

[gemredpanda]

Which of the following has a unique solution on the whole interval (0, pi)?
y''+y=0, y(0)=0, y(pi)=0
y''+4y=0, y'(0)=0, y'(pi)=0
(t+1)y''+ty=0, y(1)=1, y'(1)=0
(t-1)y'+2y=0, y(0)=0, y'(0)=1

I'm not sure where to go on this one. I solved the first 2 equations and got:
y=c*sin(t)
and y=c*cost(2t) respectively
I don't believe these are unique solutions due to the arbitrary constants in both equations.

I'm not sure I was even taught how to solve the third one. So I'm left with the 4th one as the answer. Am I correct?

 

[gemredpanda]

Actually there is no solution for the fourth since we get y=0, and y'(0)=1 wouldn't make sense...is the third one the answer?

 

[LCKurtz]

Actually there is no solution for the fourth since we get y=0, and y'(0)=1 wouldn't make sense...is the third one the answer?

You've correctly ruled out a and b.

There's nothing wrong with specifying y(0)=0 and y'(0) = 1. But I think you are supposed to think about what your existence and uniqueness theorem for initial value problems says.

 

[gemredpanda]

I didn't mean that it was wrong to specify y(0)=0 and y'(0)=1, I meant that if we solve the 4th equatio using the integrating factor method we would get

μ=e^(∫(2 / (t-1) ) )
which reduces to μ=(t-1)^2

then we get
y[(t-1)^2]= ∫0
so we get y[(t-1)^2]=c

invoking y(0)=0 we get
so we get y=0.

Then if we try to invoke y'(0)=1 we can't, since that makes no sense. Am I right? Thanks for you help!

EDIT: Oh, I think I got it...
the fourth equation has a discontinuity at t=1, so the intervals are (-∞,1) (1,∞)---the original IC have the point t=0, so the largest interval on which a unique solution is guaranteed is (-∞,1). So this isn't our answer,

the third equation has a discontinutity at t=-1. So the intervals are (-∞,-1) and (-1,∞)
Here the ICs have the point t=1, which is in the interval (-1,∞). So (-1,∞) is the largest interval on which a unique solution is guananteed, so this must be the answer.

I'm also having troubles on some superposition problems if anyone can help...
Given Y1 is a solution to y''+py'+qy=e^t
and Y2 is a solution to y''+py'+qy=3e^t
Which of the following is a solution to y''+py'+qy=2e^t?
Y=Y1+Y2
Y=Y1-Y2
Y=2Y1
Y=2Y2

I know that given the Y1 and Y2 where Y1 and Y2 are the solutions to a homogenous diff eq we know that
Y3=c1Y1+c2Y2
is also a solution. But I can't seem to reconcile how to apply this here. my guess at the answer would just be Y=2Y1, since 2e^t is 2 times e^t...thats probably way wrong thoughAnd another similar superpostition problems is:

Y1 and Y2 are solutions to y''+py'+qy=g(t). Which one of the following is also a solution
Y=Y1-Y2
Y=Y1+Y2
Y=2Y1-Y2
Y=2Y1-3Y2
I think this is also a superposition question but I'm unsure of what to do.

 

[HallsofIvy]

Y1''+ pY1'+ qY1= 3e^t and Y2''+ pY2'+ qY2= 2e^t so if
Y= Y1+ Y2, Y''+ pY'+ qY= (Y1+ Y2)''+ p(Y1+ Y2)'+ q(Y1+ Y2)= Y1''+ Y2''+ pY1'+ pY2'+ qY1+ qY2= (Y1''+ pY1'+ qY1)+ (Y2''+ pY2'+ qY2)= (2e^t)+ (2e^t)= 4e^t, not 2e^t so Y1+ Y2 does NOT satisfy the equation., Do the same with the others.