Differential equation

[God]

Okay so I have two assertions, I have to demonstrate whether they're true or false. i'd like help on the first one especially.
I think I found a proof for the second one, but i'd like someone to check whether what I say is true or not...
thanks in advance

First one : If y'(x)=sin(y²(x)-x²)+1/2 and y(0)=-1 then x>y(x) for x>=0

Of course, solving such an equation isn't the way to go ( I don't have enough knowledge to do so anyway )

I thought about two alternatives, but none of them seem to work

first alternative was using Cauchy's theorem which states two solutions don't cross on a graph for such an equation ( finding an obvious root of this equation with a function superior to -1 at x=0, inferior or equal to f(x)=x on R+ and defined on R+ would have worked )
but unfortunately, y(x)=x isn't a solution of this equation and I couldn't find any obvious root so I can't use it

2nd alternative: proving that it cannot be superior to x using the variations of y'
I know sin is between -1 and 1, but y'=sin( )+1/2 so -1/2<y'<3/2 so y' can be superior to 1, so it means y can grow "faster" than f(x)=x since x'=1
so I can't prove the graph of the solution will stay under y(x)=x...

I have really no idea on how to work this one out...


2nd question : y'=-sin(y), if -pi/2<y(0)<pi/2 then -pi/2<f(x)<pi/2 for x in R

What I thought :

if -pi/2<y(0)<0 then y'=-sin(y) >0 until y reaches 0, when it is reached y stays constant equal to 0 since y'=sin(0)=0 so it works (f(x) stays between -pi/2 and 0)
if y(0)=0 then y is constant since y'=-sin(0)=0 ... so it works too
if 0<y<pi/2 then y'=-sin(y)<0 and stays negative until y hits 0, and when it does, y'=0 so it stays constant and it works

so i'd say that one is true...

 

[girodd]

For your first question:

Let a be the smallest positive number such that y(b)=b. We would like to show that y(x)<x for all positive x, i.e., that b does not exist. Since b is positive, for any positive a<b we have y(a) < a, and so (y(b) - y(a)) / (b-a) = (b-y(a))/(b-a) > 1, and by the mean value theorem there must be a z between a and b with y'(z) > 1. However, for all x such that y(x) is close to x we have y'(x) close to 1/2..... You can work this out with epsilons and deltas and prove that b cannot exist. Something similar will work in your second question as well.

 

[God]

For your first question:

Let a be the smallest positive number such that y(b)=b. We would like to show that y(x)<x for all positive x, i.e., that b does not exist. Since b is positive, for any positive a<b we have y(a) < a, and so (y(b) - y(a)) / (b-a) = (b-y(a))/(b-a) > 1, and by the mean value theorem there must be a z between a and b with y'(z) > 1. However, for all x such that y(x) is close to x we have y'(x) close to 1/2..... You can work this out with epsilons and deltas and prove that b cannot exist. Something similar will work in your second question as well.

>Let a be the smallest positive number such that y(b)=b.

I suppose you mean b and not a


well I have no idea how I can work this out with epsilons and deltas although I have heard about it, I guess it's well above my level ( I'm only in my first college year )
all I know is the mean value theorem.
I think there is an easier way to prove it
thanks a lot though