Q. B - Calculus

Question

A certain set of parametric equations has the properties that:

$x\text{'} \left(t\right) = x\left(t\right), y\text{'}\left(t\right) = x\left(t\right)y\left(t\right), x\left(0\right)=1 \mathrm{and} y\left(0\right) = e$

(a) Solve the system of differential equations. (Hint: Start by solving the initial-value problem $x\text{'}\left(t\right) = x\left(t\right) \mathrm{with} x\left(0\right) = 1$)

(b) Graph the parametric curve defined by $x = x\left(t\right) \mathrm{and} y = y\left(t\right)$by eliminating the parameter 

Step-by-Step Solution

Answer

(a).

(b). The graph is,

Step 1: Given information

$x\text{'} \left(t\right) = x\left(t\right), y\text{'}\left(t\right) = x\left(t\right)y\left(t\right), x\left(0\right)=1 \mathrm{and} y\left(0\right) = e$

Part (a) Step 1: Solve

We have,

$$\begin{gathered} x\text{'} \left(t\right) = x\left(t\right) \\ \Rightarrow \int x\text{'} \left(t\right) dt =\int x\left(t\right) dt \\ \Rightarrow x \left(0\right)=\int x\left(0\right) dt+c \\ \Rightarrow 1 =\int 1 dt+c \\ \Rightarrow 1 =t+c \\ y\text{'}\left(t\right) = x\left(t\right)y\left(t\right) \\ \Rightarrow \int y\text{'}\left(t\right) dt= \int x\left(t\right)y\left(t\right) dt \\ \Rightarrow \int y\text{'}\left(t\right) dt= x\left(t\right)\int y\left(t\right) dt-\int \left( x\text{'}\left(t\right)\int y\left(t\right) dt\right) dt \\ \Rightarrow y\left(t\right) = x\left(t\right)\int y\left(t\right) dt-\int \left( x\text{'}\left(t\right)\int y\left(t\right) dt\right) dt \\ \Rightarrow y\left(0\right) = x\left(0\right)\int y\left(0\right) dt-\int \left( x\text{'}\left(0\right)\int y\left(0\right) dt\right) dt \\ \Rightarrow e =1\int edt-\int \left(1\int e dt\right) dt \\ \Rightarrow e =et-\frac{e{t}^2}{2}+c \end{gathered}$$

Part (b) Step 1: Graph

The graph of the function is as shown below;