Existence and Uniqueness of the Solution to the ODE

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

Theorem 1. Consider the nth-order linear differential equation ${{y}}^{{{\left({n}\right)}}}+{p}_{{1}}{\left({t}\right)}{{y}}^{{{\left({n}-{1}\right)}}}+{p}_{{2}}{\left({t}\right)}{{y}}^{{{\left({n}-{2}\right)}}}+\ldots+{p}_{{n}}{\left({t}\right)}={f{{\left({t}\right)}}}$. If all coefficients ${p}_{{1}}{\left({t}\right)}$, ${p}_{{2}}{\left({t}\right)}$, ..., ${p}_{{n}}{\left({t}\right)}$ and ${f{{\left({t}\right)}}}$ are continuous on the interval ${\left({a},{b}\right)}$, the equation has the unique solution which satisfies the given initial conditions ${y}{\left({t}_{{0}}\right)}={y}_{{0}}$, ${y}'{\left({t}_{{0}}\right)}={{y}_{{0}}^{'}}$, ..., ${{y}}^{{{\left({n}-{1}\right)}}}{\left({t}_{{0}}\right)}={{y}_{{0}}^{{{\left({n}-{1}\right)}}}}$, where ${t}_{{0}}$ belongs to the interval ${\left({a},{b}\right)}$.

Note that there are no restrictions on ${y}_{{0}},{{y}_{{0}}^{'}},\ldots,{{y}_{{0}}^{{{\left({n}-{1}\right)}}}}$.

Theorem 2. Consider the IVP ${y}'={f{{\left({t},{y}\right)}}}$ ${y}{\left({t}_{{0}}\right)}={y}_{{0}}$. If ${f{{\left({t},{y}\right)}}}$ and $\frac{{\partial{f}}}{{\partial{y}}}$ are continuous in some rectangle ${\left({t}_{{1}},{t}_{{2}}\right)}$, ${\left({y}_{{1}},{y}_{{2}}\right)}$ that contains the point ${\left({t}_{{0}},{y}_{{0}}\right)}$, there exists a unique solution to the IVP on some interval ${t}_{{0}}-\epsilon<{t}<{t}_{{0}}+\epsilon$ $\left(\epsilon>{0}\right)$ that is contained in the interval ${\left({t}_{{1}},{t}_{{2}}\right)}$.

These theorems are very similar, however there is one big difference: for the linear ODE (theorem 1), there are no restrictions on the initial function values $\left({y}_{{0}},{{y}_{{0}}^{'}},\ldots,{{y}_{{0}}^{{{\left({n}-{1}\right)}}}}\right)$, unlike for the non-linear ODE (theorem 2).

Example. Determine all the solutions to the following IVP: ${y}'={{y}}^{{\frac{{1}}{{5}}}}$, ${y}{\left({0}\right)}={0}$.

Here, ${f{{\left({t},{y}\right)}}}={{y}}^{{\frac{{1}}{{5}}}}$ and it is continuous, but $\frac{{\partial{f}}}{{\partial{y}}}=\frac{{1}}{{5}}{{y}}^{{-\frac{{4}}{{5}}}}$ 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 $\frac{{{d}{y}}}{{{d}{t}}}={{y}}^{{\frac{{1}}{{5}}}}\to\frac{{{d}{y}}}{{{{y}}^{{\frac{{1}}{{5}}}}}}={d}{t}$.

Integrating both sides gives $\frac{{5}}{{4}}{{y}}^{{\frac{{4}}{{5}}}}={t}+{C}$.

Plugging the initial conditions yields $\frac{{5}}{{4}}\cdot{{0}}^{{\frac{{4}}{{5}}}}={0}+{C}\to{C}={0}$.

So, $\frac{{5}}{{4}}{{y}}^{{\frac{{4}}{{5}}}}={t}$,

Or

${{y}}^{{\frac{{4}}{{5}}}}=\frac{{4}}{{5}}{t}$

${{y}}^{{4}}={{\left(\frac{{4}}{{5}}{t}\right)}}^{{5}}$

${y}=\pm{{\left(\frac{{4}}{{5}}{t}\right)}}^{{\frac{{5}}{{4}}}}$

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