Question

Please answer the following questions that a person new to this course would be able to understand.( Include theorem.)

Problem 6: Consider the linear systems of differential equations a) Sketch the direction Seld for the line gystem. write StreamPlotl(x-2y, 2x-3y] İn Wolframı Alpha a) Use the method of elimination to find a second order linear differential equation that is satisfied by (t b) Find particular solutions x(t) and y(t) such that x(0) 1 and y(0) 2

Answer 1

Given the linear systems of differential equations
   
Given the linear systems of differential equations x' = v_x = x - 2y y' = v_y = 2x - 3y The sketch of the direction field (v_x, v_y) in this case is We use the method of elimination by taking one more time derivative of the 1st equation of x(t) x'' = x' - 2y' Now in order to eliminate y'(t), we use the 2nd equation x'' = x' - 2(2x - 3y) And so, x'' = x' - 4x + 6y And from the first equation, we replace y, i.e., x' = x - 2y rightdoublearrow 2y = x - x' And this implies x'' = x' - 4x + 3(x - x') x'' = -2x' - x x'' + 2x' + x = 0 - (1) This is the second order differential equation. And let us assume that the ansatz for the solution is x(t) = Ae^at And so, putting this in the equation (1), we get A(alpha^2 + 2 alpha + 1) e^alpha t = 0

The sketch of the direction field (v_x, v_y) in this case is
  

a)

   We use the method of elimination by taking one more time derivative of the 1st equation of x(t)
     
Now in order to eliminate y'(t), we use the 2nd equation
   
And so,
  
And from the first equation, we replace y, i.e.,
    $x'=x-2y\Rightarrow 2y=x-x'$
And this implies

   $\Rightarrow x''= -2x'-x$
    $\Rightarrow x''+2x'+x=0$ ................(1)
This is the second order differential equation.

b)

  And let us assume that the ansatz for the solution is
   $x(t)=Ae^{\alpha t}$
And so, putting this in the equation (1), we get
   $A(\alpha^2+2\alpha +1)e^{\alpha t}=0$
   $\Rightarrow \alpha^2+2\alpha +1=0$
    $\Rightarrow \alpha=-1,~-1$
So, the two roots are equal. And so, the general solution is
    $x(t)=(A+Bt)e^{\alpha t}=(A+Bt)e^{-t}$
where, A and B are two undetermined constants which are to be fixed by the initial conditions.

  The initial conditions are
   $x(0)=1$
And
    $x'(0)=x(0)-2y(0)=1-(2\times 2)=-3$
   $\Rightarrow x'(0)=-3$

The first condition $x(0)=1$ implies
   $x(0)=A\Rightarrow A=x(0)=1$
And taking the derivative of the solution x(t), we get

   $\Rightarrow x'(t)=\left (B-(A+Bt) \right )e^{-t}$
And so, the 2nd initial condition implies
    $x'(0)=\left (B-A \right )$
Now given
   $\Rightarrow x'(0)=-3$
So, we get
   
Nw we have already got, A = 1, so, we get
   
   $\Rightarrow B =-2$
And so, the solution is
$x(t)=(1-2t)e^{-t}$
And so,
    $x'(t)=-(1-2t)e^{-t}-2e^{-t}$
   $\Rightarrow x'(t)=-(3-2t)e^{-t}$
And so, from the first equation,
    $x'(t)=x(t)-2y(t)$
   $\Rightarrow y(t)=\frac{1}{2}\left (x(t)-x'(t) \right )$
So, as we have already got an answer for x(t) and x'(t), so, we get
    $\Rightarrow y(t)=\frac{1}{2}\left ((1-2t)e^{-t}+(3-2t)e^{-t}\right )$
   $\Rightarrow y(t)=\frac{1}{2}\left (1-2t+3-2t\right )e^{-t}$
   $\Rightarrow y(t)=\frac{1}{2}\left (4-4t\right )e^{-t}$
   $\Rightarrow y(t)=2\left (1-t\right )e^{-t}$
This is the solution for y(t).