Writing a Second-Order Equation as a System of First Order Equations Write the equation [latex]\bar{y}^{\prime \prime}=0.1\left(y^{\prime}\right)^2-\frac{1600}{(40+y)^2}[/latex] as a system of first-order equations. Then, find any equilibrium solutions and interpret the result.

The idea is to define new functions u and v where u = y and v = y′. We then have u′ = y′= v and [latex]v^{\prime}=y^{\prime \prime}=0.1\left(y^{\prime}\right)^2-\frac{1600}{(40+y)^2}=0.1 v^2-\frac{1600}{(40+u)^2} .[/latex] Summarizing, we have the system of first-order equations u′= v [latex]v^{\prime}=0.1 v^2-\frac{1600}{(40+u)^2}[/latex]. The equilibrium points are then solutions of 0 = v [latex]0=0.1 v^2-\frac{1600}{(40+u)^2}[/latex]. With v = 0, observe that the second equation has no solution, so that there are no equilibrium points. For a falling object, the position (u) is not constant and so, there are no equilibrium solutions.

Question 7.4.4: Writing a Second-Order Equation as a System of First Order E...

Calculus: Early Transcendental Functions

Writing a Second-Order Equation as a System of First Order Equations

Write the equation $\bar{y}^{\prime \prime}=0.1\left(y^{\prime}\right)^2-\frac{1600}{(40+y)^2}$ as a system of first-order equations. Then, find any equilibrium solutions and interpret the result.

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The idea is to define new functions u and v where u = y and v = y′. We
then have u′ = y′= v and

v^{\prime}=y^{\prime \prime}=0.1\left(y^{\prime}\right)^2-\frac{1600}{(40+y)^2}=0.1 v^2-\frac{1600}{(40+u)^2} .

Summarizing, we have the system of first-order equations

u′= v

$v^{\prime}=0.1 v^2-\frac{1600}{(40+u)^2}$ .

The equilibrium points are then solutions of

0 = v

$0=0.1 v^2-\frac{1600}{(40+u)^2}$ .

With v = 0, observe that the second equation has no solution, so that there are no equilibrium points. For a falling object, the position (u) is not constant and so, there are no equilibrium solutions.