1.2 Initial Value Problems

1.2 Definitions

IVP:  An ODE with appropriate initial conditions

1st Order IVP: y'=f(x,y) y((Subscript[x, 0]))=(Subscript[y, 0]);

2nd Order IVP: y''=f(x,y,y'), y((Subscript[x, 0]))=(Subscript[y, 0]) and y'((Subscript[x, 0]))=(Subscript[y, 1])

Point of this is Thm 1.2.1 
	"There is one and only one solution for a 1st order IVP for nice functions f!"

For nasty functions f there could be more than one solution to a 1st order IVP.

1.1 Examples

Ex1

A 1st order IVP with solution and no constants.

TMax=12;
IVP1 =  {
y'[t]==t y[t],
y[0]==1
}
DSolve[IVP1,y[t],t]

Not all such IVPs have formulas for their solution but we can easily compute them numerically.

ySol[t_]=y[t]/.NDSolve[
IVP1,
y[t],
{t, 0, TMax}
][[1]]
Plot[
ySol[t],{t, 0, TMax}
]

Ex2

A 2nd order DE with two parameters in the solution

TMax=12;
DE2 =  {
y''[t]==y[t]
}
DSolve[DE2,y[t],t]

An IVP for this with solution and plot

IVP2 =  {
y''[t]==y[t],
y[0]==y0,
y'[0]==y1
}
DSolve[IVP2,y[t],t]

Ex3

The numerical solution of a second order IVP

TMax=120;
IVP2 =  {
y''[t]==Sin[y[t]],
y[0]==0.3,
y'[0]==0.1
}
ySol[t_]=y[t]/.NDSolve[IVP2,y[t],{t,0, TMax}][[1]]
Plot[ySol[t],{t, 0, TMax}]