Q5.4-2E - 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.

$$\begin{gathered} \frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{1}\mathbf{,}\frac{\mathbf{dy}}{\mathbf{dt}}\mathbf{=}\mathbf{3}{\mathbf{x}}^{\mathbf{2}}\mathbf{;} \\ \mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{t}\mathbf{+}\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{3}{\mathbf{t}}^{\mathbf{2}}\mathbf{+}\mathbf{3}\mathbf{t} \end{gathered}$$

Step-by-Step Solution

Answer

Hence, it is verified that $\mathrm{x}\left(\mathrm{t}\right),\mathrm{y}\left(\mathrm{t}\right)$ is a solution to the given system.

Get the result in form of x and y

Here the system is;

$$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dt}}=1 \\ \frac{\mathrm{dy}}{\mathrm{dt}}=3{\mathrm{x}}^2 \end{gathered}$$

 

And

 $$\begin{gathered} \mathrm{x}\left(\mathrm{t}\right)=\mathrm{t}+1 \\ \mathrm{y}\left(\mathrm{t}\right)={\mathrm{t}}^3+3{\mathrm{t}}^2+3\mathrm{t} \end{gathered}$$

 

Then

 $$\begin{gathered} \frac{\mathrm{dx}}{\mathrm{dt}}=1 \\ \frac{\mathrm{dy}}{\mathrm{dt}}=3({\mathrm{t}}^2+2\mathrm{t}+1)=3(\mathrm{t}+1{)}^2=3\left(\mathrm{x}\right(\mathrm{t}){)}^2 \end{gathered}$$

Therefore, the pair (x(t), y(t)) is a solution to the system.

Also,

$$\begin{gathered} \mathrm{y}\left(\mathrm{t}\right)+1={\mathrm{t}}^3+3{\mathrm{t}}^2+3\mathrm{t}+1=(\mathrm{t}+1{)}^3 \\ =\left(\mathrm{x}\right(\mathrm{t}){)}^3 \\ \mathrm{y}={\mathrm{x}}^3-1 \end{gathered}$$

Get the result

Since$\frac{\mathbf{dx}}{\mathbf{dt}}\mathbf{=}\mathbf{1}$x is increasing, the flow is from left to right along the curve.

Sketch

This is the required result.