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

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

**Answer:**

\(F(t,y)= 6 \, {\left(3 \, {y} + 4 \, t + 38\right)}^{\frac{1}{3}} \) 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.

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

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

**Answer:**

\(F(t,y)= 3 \, {\left(2 \, {y} + 4 \, t - 24\right)}^{\frac{7}{5}} \) 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.

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

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

**Answer:**

\(F(t,y)= 2 \, {\left(6 \, t + 3 \, {y} - 36\right)}^{\frac{8}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 3 \, {\left(2 \, {y} - 4 \, t - 2\right)}^{\frac{1}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 6 \, {\left(2 \, {y} - 4 \, t + 10\right)}^{\frac{1}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 3 \, {\left(3 \, {y} + 6 \, t + 27\right)}^{\frac{1}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 6 \, {\left(5 \, t + 2 \, {y} - 14\right)}^{\frac{8}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 3 \, {\left(3 \, {y} + 3 \, t - 27\right)}^{\frac{7}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 6 \, {\left(2 \, {y} - 2 \, t + 18\right)}^{\frac{2}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 6 \, {\left(2 \, {y} - 4 \, t + 4\right)}^{\frac{1}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(3 \, {y} - 2 \, t + 25\right)}^{\frac{1}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 4 \, {\left(3 \, {y} + 6 \, t\right)}^{\frac{1}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 3 \, {\left(3 \, {y} - 6 \, t + 18\right)}^{\frac{1}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(3 \, {y} - 6 \, t - 18\right)}^{\frac{8}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(3 \, {y} - 6 \, t - 6\right)}^{\frac{8}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 4 \, {\left(2 \, {y} + 3 \, t - 2\right)}^{\frac{4}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 6 \, {\left(2 \, {y} - 3 \, t - 6\right)}^{\frac{2}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 6 \, {\left(3 \, {y} - 4 \, t + 30\right)}^{\frac{4}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(2 \, {y} - 3 \, t - 27\right)}^{\frac{1}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 5 \, {\left(2 \, t + 2 \, {y} + 14\right)}^{\frac{2}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 5 \, {\left(-6 \, t + 3 \, {y} + 42\right)}^{\frac{2}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 5 \, {\left(3 \, {y} + 2 \, t + 14\right)}^{\frac{1}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(5 \, t + 2 \, {y} - 19\right)}^{\frac{2}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 5 \, {\left(2 \, {y} - 4 \, t + 2\right)}^{\frac{8}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 4 \, {\left(-4 \, t + 3 \, {y} - 25\right)}^{\frac{4}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(3 \, {y} + 3 \, t + 33\right)}^{\frac{2}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 5 \, {\left(2 \, {y} + 5 \, t - 23\right)}^{\frac{7}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 4 \, {\left(-2 \, t + 2 \, {y} - 2\right)}^{\frac{4}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(3 \, {y} + 4 \, t + 18\right)}^{\frac{1}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(-2 \, t + 3 \, {y}\right)}^{\frac{7}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 6 \, {\left(2 \, {y} + 4 \, t + 10\right)}^{\frac{8}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(2 \, {y} - 4 \, t + 20\right)}^{\frac{7}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 5 \, {\left(3 \, {y} + 3 \, t + 27\right)}^{\frac{8}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 6 \, {\left(2 \, {y} + 3 \, t - 19\right)}^{\frac{1}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 4 \, {\left(2 \, {y} - 2 \, t + 8\right)}^{\frac{2}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 3 \, {\left(3 \, {y} + 4 \, t - 42\right)}^{\frac{8}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 4 \, {\left(3 \, {y} - 4 \, t + 26\right)}^{\frac{2}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 5 \, {\left(3 \, {y} + 5 \, t + 48\right)}^{\frac{8}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 6 \, {\left(-3 \, t + 2 \, {y} + 22\right)}^{\frac{1}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(3 \, {y} - 6 \, t + 30\right)}^{\frac{7}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 4 \, {\left(2 \, t + 3 \, {y} - 7\right)}^{\frac{1}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 5 \, {\left(3 \, {y} + 3 \, t - 18\right)}^{\frac{7}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 4 \, {\left(3 \, {y} - 5 \, t - 42\right)}^{\frac{8}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 4 \, {\left(2 \, {y} + 6 \, t + 8\right)}^{\frac{4}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(3 \, {y} + 6 \, t + 30\right)}^{\frac{7}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 3 \, {\left(3 \, {y} - 2 \, t - 13\right)}^{\frac{4}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 5 \, {\left(3 \, {y} + 2 \, t + 1\right)}^{\frac{2}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(2 \, {y} + 5 \, t + 14\right)}^{\frac{8}{3}} \) 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.

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

**Answer:**

\(F(t,y)= 5 \, {\left(2 \, {y} - 6 \, t - 16\right)}^{\frac{1}{5}} \) 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.

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

**Answer:**

\(F(t,y)= 2 \, {\left(-5 \, t + 2 \, {y} + 16\right)}^{\frac{8}{5}} \) 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.