Q5.4-1E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problems 1 and 2, verify that the pair x(t), and y(t) is a solution to the given system. Sketch the trajectory of the given solution in the phase plane.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{3}{\mathbf{y}}^{\mathbf{3}}\mathbf{,}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{y}\mathbf{;} \\ \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{3}\mathbf{t}}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{e}}^{\mathbf{t}} \end{gathered}$$

Step-by-Step Solution

Answer

By putting the values of $\mathrm{x}\left(\mathrm{t}\right), \mathrm{y}\left(\mathrm{t}\right)$, get the result.

Get the result in form of x and y

Here the system is 

$$\begin{gathered} \frac{dx}{dt}=3y^3 \\ \frac{dy}{dt}=y \end{gathered}$$

And

$$\begin{gathered} x\left(t\right)=e^{3t} \\ y\left(t\right)=e^t \end{gathered}$$

Then

$$\begin{gathered} \frac{dx}{dt}=3e^{3t} \\ =3{\left(y\left(t\right)\right)}^3 \\ \frac{dy}{dt}=e^t \\ =y\left(t\right) \end{gathered}$$

Therefore, the pair is a solution to the system.

Also, ${\left(y\left(t\right)\right)}^3=x\left(t\right)=x^{\frac{1}{3}}$

Get the result

Since $\frac{dx}{dt}=3y^3$ x is increasing when $\mathbf{y}\mathbf{>}\mathbf{0}$ and x is decreasing for $\mathit{y}\mathbf{<}\mathbf{0}$. This means the flow is from left to right along the part of the curve that lies above the x-axis, and the flow is from right to left along the part of the curve that lies below the x-axis.

Sketch the graph.

This is the required result.