# 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.