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**Problem:**"Consider the initial value problem [tex]y^{\prime}=y^{1/3}, \mbox{ }y(0)=0[/tex] from Example 3 in the text.

(a) Is there a solution that passes through the point (1,1)? If so, find it.

(b) Is there a solution that passes through the point (2,1)? If so, find it.

(c) Consider all possible solutions of the given initial value problem. Determine the set of values that these solutions have at [tex]t=2[/tex]."

**Example 3:**[tex]y^{\prime}=y^{1/3}, \qquad y(0)=0 \quad \qquad (1)[/tex]

[tex]y^{-1/3} \: dy = dt[/tex]

[tex]\int y^{-1/3} \: dy = \int dt[/tex]

[tex]\frac{3}{2} y^{2/3} = t + \mathrm{C}[/tex]

[tex]y = \left[ \frac{2}{3} \left( t + \mathrm{C} \right) \right] ^{3/2}[/tex]

The initial condition is satisfied if [tex]\mathrm{C}=0[/tex], so

[tex]y=\phi _1 (t) = \left( \frac{2}{3}t \right) ^{3/2}, \qquad t \geq 0 \quad \qquad (2)[/tex]

On the other hand, the function

[tex]y=\phi _2 (t) = -\left( \frac{2}{3}t \right) ^{3/2}, \qquad t \geq 0 \quad \qquad (3)[/tex]

is yet another solution. For an arbitrary positive [tex]t_0[/tex] the functions

[tex]y=\left\{ \begin{array}{ll} 0, &\mbox{ if } 0 \leq t < t_0 \\ \pm \left[ \frac{2}{3} \left( t - t_0 \right) \right] ^{3/2} , &\mbox{ if } t \geq t_0 \end{array} \right. \quad \qquad (4)[/tex]

are solutions of Eq. (1).

**Answers:**"

**(a)**No.

**(b)**Yes; set [tex]t_0=\frac{1}{2}[/tex] in Eq. (4) from Example 3

**(c)**[tex]\left| y \right| \leq \left( \frac{4}{3} \right) ^{3/2}[/tex]"

**My work:**(a) (Why is this wrong?)

[tex]y^{\prime}=y^{1/3}, \qquad y\left( 1 \right)=1[/tex]

[tex]y^{-1/3} \: dy = dt[/tex]

[tex]\int y^{-1/3} \: dy = \int dt[/tex]

[tex]\frac{3}{2} y^{2/3} = t + \mathrm{C}[/tex]

[tex]y = \left[ \frac{2}{3} \left( t + \mathrm{C} \right) \right] ^{3/2}[/tex]

The initial condition is satisfied if [tex]\mathrm{C}=\frac{1}{2}[/tex], so

[tex]y= \left[ \frac{2}{3} \left( t + \frac{1}{2} \right) \right] ^{3/2}[/tex]

which seems to work.

(b) (This one seems to be ok.)

[tex]y^{\prime}=y^{1/3}, \qquad y\left( 2 \right)=1[/tex]

[tex]y^{-1/3} \: dy = dt[/tex]

[tex]\int y^{-1/3} \: dy = \int dt[/tex]

[tex]\frac{3}{2} y^{2/3} = t + \mathrm{C}[/tex]

[tex]y = \left[ \frac{2}{3} \left( t + \mathrm{C} \right) \right] ^{3/2}[/tex]

The initial condition is satisfied if [tex]\mathrm{C}=-\frac{1}{2}[/tex], so

[tex]y= \left[ \frac{2}{3} \left( t - \frac{1}{2} \right) \right] ^{3/2}[/tex]

(c) (This one seems to be ok.)

Considering all possible solutions of the given initial value problem, the maximum value at [tex]t=2[/tex] is given by the solution

[tex]y = \left( \frac{2}{3}t \right) ^{3/2} \Rightarrow y(2) = \left( \frac{4}{3} \right) ^{3/2}[/tex]