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...
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...