## X3 - Existence/uniqueness theorem for first-order IVPs

#### Example 1 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 6 \, {\left(3 \, {y} + 4 \, t + 38\right)}^{\frac{1}{3}} \hspace{2em} x( -5 )= -6$

$$F(t,y)= 6 \, {\left(3 \, {y} + 4 \, t + 38\right)}^{\frac{1}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{6}{{\left(3 \, {y} + 4 \, t + 38\right)}^{\frac{2}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 2 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 3 \, {\left(2 \, {y} + 4 \, t - 24\right)}^{\frac{7}{5}} \hspace{2em} x( 3 )= 6$

$$F(t,y)= 3 \, {\left(2 \, {y} + 4 \, t - 24\right)}^{\frac{7}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{42}{5} \, {\left(2 \, {y} + 4 \, t - 24\right)}^{\frac{2}{5}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 3 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(6 \, t + 3 \, {y} - 36\right)}^{\frac{8}{3}} \hspace{2em} x( 3 )= 6$

$$F(t,y)= 2 \, {\left(6 \, t + 3 \, {y} - 36\right)}^{\frac{8}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= 16 \, {\left(6 \, t + 3 \, {y} - 36\right)}^{\frac{5}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 4 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 3 \, {\left(2 \, {y} - 4 \, t - 2\right)}^{\frac{1}{3}} \hspace{2em} x( 2 )= 5$

$$F(t,y)= 3 \, {\left(2 \, {y} - 4 \, t - 2\right)}^{\frac{1}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{2}{{\left(2 \, {y} - 4 \, t - 2\right)}^{\frac{2}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 5 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 6 \, {\left(2 \, {y} - 4 \, t + 10\right)}^{\frac{1}{3}} \hspace{2em} x( 4 )= 3$

$$F(t,y)= 6 \, {\left(2 \, {y} - 4 \, t + 10\right)}^{\frac{1}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{4}{{\left(2 \, {y} - 4 \, t + 10\right)}^{\frac{2}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 6 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 3 \, {\left(3 \, {y} + 6 \, t + 27\right)}^{\frac{1}{5}} \hspace{2em} x( -6 )= 3$

$$F(t,y)= 3 \, {\left(3 \, {y} + 6 \, t + 27\right)}^{\frac{1}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{9}{5 \, {\left(3 \, {y} + 6 \, t + 27\right)}^{\frac{4}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 7 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 6 \, {\left(5 \, t + 2 \, {y} - 14\right)}^{\frac{8}{5}} \hspace{2em} x( 2 )= 2$

$$F(t,y)= 6 \, {\left(5 \, t + 2 \, {y} - 14\right)}^{\frac{8}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{96}{5} \, {\left(5 \, t + 2 \, {y} - 14\right)}^{\frac{3}{5}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 8 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 3 \, {\left(3 \, {y} + 3 \, t - 27\right)}^{\frac{7}{5}} \hspace{2em} x( 4 )= 5$

$$F(t,y)= 3 \, {\left(3 \, {y} + 3 \, t - 27\right)}^{\frac{7}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{63}{5} \, {\left(3 \, {y} + 3 \, t - 27\right)}^{\frac{2}{5}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 9 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 6 \, {\left(2 \, {y} - 2 \, t + 18\right)}^{\frac{2}{3}} \hspace{2em} x( 6 )= -3$

$$F(t,y)= 6 \, {\left(2 \, {y} - 2 \, t + 18\right)}^{\frac{2}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{8}{{\left(2 \, {y} - 2 \, t + 18\right)}^{\frac{1}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 10 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 6 \, {\left(2 \, {y} - 4 \, t + 4\right)}^{\frac{1}{3}} \hspace{2em} x( -2 )= -6$

$$F(t,y)= 6 \, {\left(2 \, {y} - 4 \, t + 4\right)}^{\frac{1}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{4}{{\left(2 \, {y} - 4 \, t + 4\right)}^{\frac{2}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 11 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(3 \, {y} - 2 \, t + 25\right)}^{\frac{1}{5}} \hspace{2em} x( 5 )= -5$

$$F(t,y)= 2 \, {\left(3 \, {y} - 2 \, t + 25\right)}^{\frac{1}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{6}{5 \, {\left(3 \, {y} - 2 \, t + 25\right)}^{\frac{4}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 12 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 4 \, {\left(3 \, {y} + 6 \, t\right)}^{\frac{1}{5}} \hspace{2em} x( 2 )= -4$

$$F(t,y)= 4 \, {\left(3 \, {y} + 6 \, t\right)}^{\frac{1}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{12}{5 \, {\left(3 \, {y} + 6 \, t\right)}^{\frac{4}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 13 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 3 \, {\left(3 \, {y} - 6 \, t + 18\right)}^{\frac{1}{5}} \hspace{2em} x( 6 )= 6$

$$F(t,y)= 3 \, {\left(3 \, {y} - 6 \, t + 18\right)}^{\frac{1}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{9}{5 \, {\left(3 \, {y} - 6 \, t + 18\right)}^{\frac{4}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 14 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(3 \, {y} - 6 \, t - 18\right)}^{\frac{8}{3}} \hspace{2em} x( -5 )= -4$

$$F(t,y)= 2 \, {\left(3 \, {y} - 6 \, t - 18\right)}^{\frac{8}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= 16 \, {\left(3 \, {y} - 6 \, t - 18\right)}^{\frac{5}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 15 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(3 \, {y} - 6 \, t - 6\right)}^{\frac{8}{5}} \hspace{2em} x( -4 )= -6$

$$F(t,y)= 2 \, {\left(3 \, {y} - 6 \, t - 6\right)}^{\frac{8}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{48}{5} \, {\left(3 \, {y} - 6 \, t - 6\right)}^{\frac{3}{5}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 16 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 4 \, {\left(2 \, {y} + 3 \, t - 2\right)}^{\frac{4}{3}} \hspace{2em} x( 2 )= -2$

$$F(t,y)= 4 \, {\left(2 \, {y} + 3 \, t - 2\right)}^{\frac{4}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{32}{3} \, {\left(2 \, {y} + 3 \, t - 2\right)}^{\frac{1}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 17 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 6 \, {\left(2 \, {y} - 3 \, t - 6\right)}^{\frac{2}{3}} \hspace{2em} x( -4 )= -3$

$$F(t,y)= 6 \, {\left(2 \, {y} - 3 \, t - 6\right)}^{\frac{2}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{8}{{\left(2 \, {y} - 3 \, t - 6\right)}^{\frac{1}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 18 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 6 \, {\left(3 \, {y} - 4 \, t + 30\right)}^{\frac{4}{3}} \hspace{2em} x( 6 )= -2$

$$F(t,y)= 6 \, {\left(3 \, {y} - 4 \, t + 30\right)}^{\frac{4}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= 24 \, {\left(3 \, {y} - 4 \, t + 30\right)}^{\frac{1}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 19 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(2 \, {y} - 3 \, t - 27\right)}^{\frac{1}{5}} \hspace{2em} x( -5 )= 6$

$$F(t,y)= 2 \, {\left(2 \, {y} - 3 \, t - 27\right)}^{\frac{1}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{4}{5 \, {\left(2 \, {y} - 3 \, t - 27\right)}^{\frac{4}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 20 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 5 \, {\left(2 \, t + 2 \, {y} + 14\right)}^{\frac{2}{3}} \hspace{2em} x( -3 )= -4$

$$F(t,y)= 5 \, {\left(2 \, t + 2 \, {y} + 14\right)}^{\frac{2}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{20}{3 \, {\left(2 \, t + 2 \, {y} + 14\right)}^{\frac{1}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 21 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 5 \, {\left(-6 \, t + 3 \, {y} + 42\right)}^{\frac{2}{5}} \hspace{2em} x( 5 )= -4$

$$F(t,y)= 5 \, {\left(-6 \, t + 3 \, {y} + 42\right)}^{\frac{2}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{6}{{\left(-6 \, t + 3 \, {y} + 42\right)}^{\frac{3}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 22 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 5 \, {\left(3 \, {y} + 2 \, t + 14\right)}^{\frac{1}{5}} \hspace{2em} x( -4 )= -2$

$$F(t,y)= 5 \, {\left(3 \, {y} + 2 \, t + 14\right)}^{\frac{1}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{3}{{\left(3 \, {y} + 2 \, t + 14\right)}^{\frac{4}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 23 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(5 \, t + 2 \, {y} - 19\right)}^{\frac{2}{3}} \hspace{2em} x( 3 )= 2$

$$F(t,y)= 2 \, {\left(5 \, t + 2 \, {y} - 19\right)}^{\frac{2}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{8}{3 \, {\left(5 \, t + 2 \, {y} - 19\right)}^{\frac{1}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 24 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 5 \, {\left(2 \, {y} - 4 \, t + 2\right)}^{\frac{8}{5}} \hspace{2em} x( 2 )= 3$

$$F(t,y)= 5 \, {\left(2 \, {y} - 4 \, t + 2\right)}^{\frac{8}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= 16 \, {\left(2 \, {y} - 4 \, t + 2\right)}^{\frac{3}{5}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 25 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 4 \, {\left(-4 \, t + 3 \, {y} - 25\right)}^{\frac{4}{5}} \hspace{2em} x( -4 )= 3$

$$F(t,y)= 4 \, {\left(-4 \, t + 3 \, {y} - 25\right)}^{\frac{4}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{48}{5 \, {\left(-4 \, t + 3 \, {y} - 25\right)}^{\frac{1}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 26 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(3 \, {y} + 3 \, t + 33\right)}^{\frac{2}{5}} \hspace{2em} x( -6 )= -5$

$$F(t,y)= 2 \, {\left(3 \, {y} + 3 \, t + 33\right)}^{\frac{2}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{12}{5 \, {\left(3 \, {y} + 3 \, t + 33\right)}^{\frac{3}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 27 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 5 \, {\left(2 \, {y} + 5 \, t - 23\right)}^{\frac{7}{3}} \hspace{2em} x( 3 )= 4$

$$F(t,y)= 5 \, {\left(2 \, {y} + 5 \, t - 23\right)}^{\frac{7}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{70}{3} \, {\left(2 \, {y} + 5 \, t - 23\right)}^{\frac{4}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 28 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 4 \, {\left(-2 \, t + 2 \, {y} - 2\right)}^{\frac{4}{5}} \hspace{2em} x( -5 )= -4$

$$F(t,y)= 4 \, {\left(-2 \, t + 2 \, {y} - 2\right)}^{\frac{4}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{32}{5 \, {\left(-2 \, t + 2 \, {y} - 2\right)}^{\frac{1}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 29 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(3 \, {y} + 4 \, t + 18\right)}^{\frac{1}{3}} \hspace{2em} x( -6 )= 2$

$$F(t,y)= 2 \, {\left(3 \, {y} + 4 \, t + 18\right)}^{\frac{1}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{2}{{\left(3 \, {y} + 4 \, t + 18\right)}^{\frac{2}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 30 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(-2 \, t + 3 \, {y}\right)}^{\frac{7}{3}} \hspace{2em} x( -6 )= -4$

$$F(t,y)= 2 \, {\left(-2 \, t + 3 \, {y}\right)}^{\frac{7}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= 14 \, {\left(-2 \, t + 3 \, {y}\right)}^{\frac{4}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 31 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 6 \, {\left(2 \, {y} + 4 \, t + 10\right)}^{\frac{8}{3}} \hspace{2em} x( -4 )= 3$

$$F(t,y)= 6 \, {\left(2 \, {y} + 4 \, t + 10\right)}^{\frac{8}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= 32 \, {\left(2 \, {y} + 4 \, t + 10\right)}^{\frac{5}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 32 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(2 \, {y} - 4 \, t + 20\right)}^{\frac{7}{3}} \hspace{2em} x( 3 )= -4$

$$F(t,y)= 2 \, {\left(2 \, {y} - 4 \, t + 20\right)}^{\frac{7}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{28}{3} \, {\left(2 \, {y} - 4 \, t + 20\right)}^{\frac{4}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 33 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 5 \, {\left(3 \, {y} + 3 \, t + 27\right)}^{\frac{8}{5}} \hspace{2em} x( -6 )= -3$

$$F(t,y)= 5 \, {\left(3 \, {y} + 3 \, t + 27\right)}^{\frac{8}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= 24 \, {\left(3 \, {y} + 3 \, t + 27\right)}^{\frac{3}{5}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 34 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 6 \, {\left(2 \, {y} + 3 \, t - 19\right)}^{\frac{1}{5}} \hspace{2em} x( 3 )= 5$

$$F(t,y)= 6 \, {\left(2 \, {y} + 3 \, t - 19\right)}^{\frac{1}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{12}{5 \, {\left(2 \, {y} + 3 \, t - 19\right)}^{\frac{4}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 35 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 4 \, {\left(2 \, {y} - 2 \, t + 8\right)}^{\frac{2}{3}} \hspace{2em} x( 2 )= -2$

$$F(t,y)= 4 \, {\left(2 \, {y} - 2 \, t + 8\right)}^{\frac{2}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{16}{3 \, {\left(2 \, {y} - 2 \, t + 8\right)}^{\frac{1}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 36 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 3 \, {\left(3 \, {y} + 4 \, t - 42\right)}^{\frac{8}{3}} \hspace{2em} x( 6 )= 6$

$$F(t,y)= 3 \, {\left(3 \, {y} + 4 \, t - 42\right)}^{\frac{8}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= 24 \, {\left(3 \, {y} + 4 \, t - 42\right)}^{\frac{5}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 37 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 4 \, {\left(3 \, {y} - 4 \, t + 26\right)}^{\frac{2}{3}} \hspace{2em} x( 5 )= -2$

$$F(t,y)= 4 \, {\left(3 \, {y} - 4 \, t + 26\right)}^{\frac{2}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{8}{{\left(3 \, {y} - 4 \, t + 26\right)}^{\frac{1}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 38 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 5 \, {\left(3 \, {y} + 5 \, t + 48\right)}^{\frac{8}{3}} \hspace{2em} x( -6 )= -6$

$$F(t,y)= 5 \, {\left(3 \, {y} + 5 \, t + 48\right)}^{\frac{8}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= 40 \, {\left(3 \, {y} + 5 \, t + 48\right)}^{\frac{5}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 39 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 6 \, {\left(-3 \, t + 2 \, {y} + 22\right)}^{\frac{1}{3}} \hspace{2em} x( 6 )= -2$

$$F(t,y)= 6 \, {\left(-3 \, t + 2 \, {y} + 22\right)}^{\frac{1}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{4}{{\left(-3 \, t + 2 \, {y} + 22\right)}^{\frac{2}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 40 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(3 \, {y} - 6 \, t + 30\right)}^{\frac{7}{5}} \hspace{2em} x( 2 )= -6$

$$F(t,y)= 2 \, {\left(3 \, {y} - 6 \, t + 30\right)}^{\frac{7}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{42}{5} \, {\left(3 \, {y} - 6 \, t + 30\right)}^{\frac{2}{5}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 41 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 4 \, {\left(2 \, t + 3 \, {y} - 7\right)}^{\frac{1}{3}} \hspace{2em} x( -4 )= 5$

$$F(t,y)= 4 \, {\left(2 \, t + 3 \, {y} - 7\right)}^{\frac{1}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{4}{{\left(2 \, t + 3 \, {y} - 7\right)}^{\frac{2}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 42 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 5 \, {\left(3 \, {y} + 3 \, t - 18\right)}^{\frac{7}{3}} \hspace{2em} x( 4 )= 2$

$$F(t,y)= 5 \, {\left(3 \, {y} + 3 \, t - 18\right)}^{\frac{7}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= 35 \, {\left(3 \, {y} + 3 \, t - 18\right)}^{\frac{4}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 43 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 4 \, {\left(3 \, {y} - 5 \, t - 42\right)}^{\frac{8}{5}} \hspace{2em} x( -6 )= 4$

$$F(t,y)= 4 \, {\left(3 \, {y} - 5 \, t - 42\right)}^{\frac{8}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{96}{5} \, {\left(3 \, {y} - 5 \, t - 42\right)}^{\frac{3}{5}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 44 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 4 \, {\left(2 \, {y} + 6 \, t + 8\right)}^{\frac{4}{3}} \hspace{2em} x( -3 )= 5$

$$F(t,y)= 4 \, {\left(2 \, {y} + 6 \, t + 8\right)}^{\frac{4}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{32}{3} \, {\left(2 \, {y} + 6 \, t + 8\right)}^{\frac{1}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 45 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(3 \, {y} + 6 \, t + 30\right)}^{\frac{7}{5}} \hspace{2em} x( -3 )= -4$

$$F(t,y)= 2 \, {\left(3 \, {y} + 6 \, t + 30\right)}^{\frac{7}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{42}{5} \, {\left(3 \, {y} + 6 \, t + 30\right)}^{\frac{2}{5}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 46 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 3 \, {\left(3 \, {y} - 2 \, t - 13\right)}^{\frac{4}{5}} \hspace{2em} x( -2 )= 3$

$$F(t,y)= 3 \, {\left(3 \, {y} - 2 \, t - 13\right)}^{\frac{4}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{36}{5 \, {\left(3 \, {y} - 2 \, t - 13\right)}^{\frac{1}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 47 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 5 \, {\left(3 \, {y} + 2 \, t + 1\right)}^{\frac{2}{3}} \hspace{2em} x( 4 )= -3$

$$F(t,y)= 5 \, {\left(3 \, {y} + 2 \, t + 1\right)}^{\frac{2}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{10}{{\left(3 \, {y} + 2 \, t + 1\right)}^{\frac{1}{3}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 48 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(2 \, {y} + 5 \, t + 14\right)}^{\frac{8}{3}} \hspace{2em} x( -4 )= 3$

$$F(t,y)= 2 \, {\left(2 \, {y} + 5 \, t + 14\right)}^{\frac{8}{3}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{32}{3} \, {\left(2 \, {y} + 5 \, t + 14\right)}^{\frac{5}{3}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.

#### Example 49 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 5 \, {\left(2 \, {y} - 6 \, t - 16\right)}^{\frac{1}{5}} \hspace{2em} x( -4 )= -4$

$$F(t,y)= 5 \, {\left(2 \, {y} - 6 \, t - 16\right)}^{\frac{1}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.

$$F_y= \frac{2}{{\left(2 \, {y} - 6 \, t - 16\right)}^{\frac{4}{5}}}$$ is not continous (or even defined) at the initial value so the guaranteed solution may not be unique.

#### Example 50 🔗

Explain what the Existence and Uniqueness Theorem for First Order IVPs guarantees about the existence and uniqueness of solutions for the following IVP.

$y'= 2 \, {\left(-5 \, t + 2 \, {y} + 16\right)}^{\frac{8}{5}} \hspace{2em} x( 2 )= -3$

$$F(t,y)= 2 \, {\left(-5 \, t + 2 \, {y} + 16\right)}^{\frac{8}{5}}$$ is continuous at and nearby the initial value so a solution exists for a nearby interval.
$$F_y= \frac{32}{5} \, {\left(-5 \, t + 2 \, {y} + 16\right)}^{\frac{3}{5}}$$ is continous at and nearby the initial value so the solution is unique for a nearby interval.