Existence and Uniqueness of the Solution the ODE

This note contains some theorems that refer to the existence and uniqueness of the solution to the ODE.

Theorem 1. Consider the n-th-order linear differential equation: `y^((n))+p_1(t)y^((n-1))+p_2(t)y^((n-2))+...+p_n(t)=f(t)`. If all coefficients `p_1(t)`, `p_2(t)`, ..., `p_n(t)` and `f(t)` are continuous on the interval `(a,b)`, the equation has the unique solution which satisfies the given initial conditions `y(t_0)=y_0`, `y'(t_0)=y_0^'`, ..., `y^((n-1))(t_0)=y_0^((n-1))`, where `t_0` belongs to the interval `(a,b)`.

Note that there are no restrictions on `y_0,y_0^',...,y_0^((n-1))`.

Theorem 2. Consider the IVP `y'=f(t,y)` `y(t_0)=y_0`. If `f(t,y)` and `(partial f)/(partial y)` are continuous in some rectangle `(t_1,t_2)`, `(y_1,y_2)` that contains the point `(t_0,y_0)`, there exists a unique solution to the IVP on some interval `t_0-epsilon<t<t_0+epsilon` (`epsilon>0`) that is contained in the interval `(t_1,t_2)`.

These theorems are very similar, however there is one big difference: for the linear ODE, there are no restrictions on the initial function values (`y_0,y_0^',...,y_0^((n-1))`), unlike for the non-linear ODE (theorem 2).

Example 1. Determine all the solutions to the following IVP: `y'=y^(1/5)`, y(0)=0.

Here, `f(t,y)=y^(1/5)` and it is continuous, but `(partial f)/(partial y)=1/5 y^(-4/5)` is not continuous at `y=0`; so, theorem 2 doesn't hold, and there is more than one solution. Note that the first solution is `y=0`.

To find the second solution, note that `(dy)/(dt)=y^(1/5)->(dy)/(y^(1/5))=dt`.

Integrating both sides gives `5/4 y^(4/5)=t+C`.

Plugging the initial conditions yields `5/4*0^(4/5)=0+C ->C=0`.

So, `5/4 y^(4/5)=t`,

Or

`y^(4/5)=4/5t`

`y^4=(4/5t)^5`

`y=+-(4/5t)^(5/4)`

This gives two other solutions. Hence, together with `y=0`, there are three different solutions.