Short Answer

In Problems 19–24, convert the given second-order equation into a first-order system by setting v=y’. Then find all the critical points in the yv-plane. Finally, sketch (by hand or software) the direction fields, and describe the stability of the critical points (i.e., compare with Figure 5.12).
$\frac{d^{2} y}{d t^{2}} + y^{3} = 0$ .

The point is the center point (0, 0).

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Step by Step Solution

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Step 1: Find the critical point

Here the equation is $\frac{d^{2} y}{d t^{2}} + y^{3} = 0$ .

Put $v = y' and v' = y''$ .

Then the system is;

$y' = v y'' = - y^{3} v' = - y^{3}$

For critical points equate the system equal to zero.

$\begin{array}{rcl} v & = & 0 \\ - y^{3} & = & 0 \\ y & = & 0 \end{array}$

So, the critical point is (0, 0).

The phase plane equation is;

$\begin{array}{rcl} \frac{d v}{d y} & = & \frac{- y}{v} \\ \int v d v & = & \int - y^{3} d y \\ \frac{v^{2}}{2} & = & - \frac{y^{4}}{4} + c \\ 4 v^{2} + y^{2} & = & c \end{array}$

Step 2: Sketch

Therefore, the point is the center point (0, 0).

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Fundamentals Of Differential Equations And Boundary Value Problems

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Step 1: Find the critical point

Step 2: Sketch

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Fundamentals Of Differential Equations And Boundary Value Problems

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Find the critical point

Step 2: Sketch

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