Q3 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Try to predict the qualitative features of the solution to $\mathrm{y}-6{\mathrm{y}}^2=0$ that satisfies the initial conditions $\mathrm{y}\left(0\right)=-1,\mathrm{y}\text{'}\left(0\right)=-1$. Compare with the computer-generated Figure 4.23. [Hint: Consider the sign of the spring stiffness]

Step-by-Step Solution

Answer

The sign of the stiffness constant is negative. And, stiffness is $\mathrm{k}=-6\mathrm{y}$. Then, the initial conditions both are negative and compared with computer-generated figure the stiffness reverses the negative velocity and restores to zero.

Step 1: General form

The mass–spring oscillator equation 

$$\begin{gathered} {\mathrm{F}}_{\mathrm{ext}}=\left[\mathrm{inertia}\right]\mathrm{y}+\left[\mathrm{damping}\right]\mathrm{y}\text{'}+\left[\mathrm{stiffness}\right]\mathrm{y} \\ =\mathrm{my}+\mathrm{by}\text{'}+\mathrm{ky} \end{gathered}$$......(1)

Step 2: Evaluate the given equation

Given that, $\mathrm{y}-6{\mathrm{y}}^2=0$ and $\mathrm{y}\left(0\right)=-1,\mathrm{y}\text{'}\left(0\right)=-1$.

 

Compare the given equation with equation (1) to get stiffness.

 

Here, the stiffness is negative $\mathrm{k}=-6\mathrm{y}$ and negative sign indicates it lends to reinforce, rather than oppose, displacement. 

 

Then, the initial conditions are both negative. So, the stiffness reverses the negative velocity and restores it to zero.