Question

Find and graph the vector function determined $\mathbf{r}\left(t\right)=⟨x\left(t\right),y\left(t\right)⟩$by the differential equation

$x^{\mathrm{\prime }}\left(t\right)=x,y^{\mathrm{\prime }}\left(t\right)=x^2,x\left(0\right)=1,y\left(0\right)=2$. (HINT: Start by solving the initial-value problem $x^{\mathrm{\prime }}\left(t\right)=x,x\left(0\right)=1$.)

Short answer

Ans: Vector function is $r\left(t\right)=<e^t,t{e}^{2t}+2>$

Graph is:

Step-by-step solution

Step 1. Given information.

given,

$x^{\mathrm{\prime }}\left(t\right)=x,y^{\mathrm{\prime }}\left(t\right)=x^2,x\left(0\right)=1,y\left(0\right)=2$

Step 2. The objective of this problem is to find and graph the vector r(t)=⟨x(t),y(t)⟩ determined by the differential equations

$\begin{array}{r}{x}^{\mathrm{\prime }}\left(t\right)=x\dots \dots \left(1\right)\\ y^{\mathrm{\prime }}\left(t\right)=x^2\dots \dots \left(2\right)\\ x\left(0\right)=1,y\left(0\right)=2\end{array}$

solve equation $1$

$x\text{'}\left(t\right)=x$

$\frac{dx}{dt}=x$

Separate the variables

$\frac{dx}{x}=dt$

Integrate both sides

$\begin{array}{r}\int \frac{dx}{x}=\int dt+c_1\\ \ln x=t+c_1\\ x=e^{t+c_1}\\ x=c{e}^t\end{array}$

Therefore,

$x=e^t.....\left(3\right)$

Step 3. Now solve 

$\begin{array}{r}{y}^{\mathrm{\prime }}\left(t\right)=x^2\\ \frac{dy}{dt}=x^2\\ dy=x^{2}dt\end{array}$

Integrate

$\begin{array}{r}\int dy=\int x^{2}dt+c_2\\ y=x^{2}t+c_2\\ \text{At}t=0,y=2\\ 2=c_2\end{array}$

Therefore,

$y=x^{2}t+2....\left(4\right)$

Use $x=e^t$from (3) into (4)

$y=t{e}^{2t}+2$

Thus, the vector of the function is

$r\left(t\right)=⟨e^t,t{e}^{2t}+2⟩$

Step 4. Draw the graph

Graph is: