MHB Solve IVP: 3.4.5.5 | Eigenvectors Found

[karush]

Solve IVP
$\begin{array}{rl}x' & = 2x + 2y\\y' & = -4x + 6y\\x(0) & = 2\\y(0) & = -3
\end{array}$
assume we can proceed with this first
$A=\left[\begin{array}{rr}2&2\\-4&6\end{array}\right]\\
A-rI=\left[\begin{array}{rr}2-r&2\\-4&6-r\end{array}\right]=r^{2} -8r + 20 = 0 \quad r_1 = 4 -2 i \quad r_2=4+2i$

eigenvector: $\left[\begin{array}{c}\dfrac{1}{2} + \dfrac{i}{2}\\1\end{array}\right]$

eigenvector: $\left[\begin{array}{c}\dfrac{1}{2} - \dfrac{i}{2}\\1\end{array}\right]$

so far

 

[HOI]

Once again, I would not use matrices for this simple a problem.

We have x'= 2x+ 2y so x''= 2x'+ 2y'.
We also have y'= -4x+ 6y so x''= 2x'+ 2(-4x+ 6y)= 2x'- 8x+ 12y.From x'= 2x+ 2y, 2y= x'- 2x so 12y= 6x'- 12x

We have x''= 2x'- 8x+ 6x'- 12x= 8x'- 20x
.x''- 8x'+ 20x= 0.

The characteristic equation is r^2- 8r+ 20= r^2- 8r+ 16- 16+ 20= (r- 4)^2+ 4= 0.
(r- 4)^2= -4 so $r- 4= \pm 2i$. $r= 4\pm 2i$

$x= e^{4t}(A cos(2t)+ B sin(2t))$
$12y= 6x'- 12x$ so $y= (1/2)x'- x$

$x'= 4e^{4t}(A cos(2t)+ B sin(2t))+ e^{4t}(-2A sin(2t)+ 2B cos(2t))= e^{4t}((4A+ 2B)cos(2t)+ (B- 2A)sin(2t))$

so $y= e^{4t}((2A+B)cos(2t)+ (B/2- A)sin(2t))- e^{4t}(A cos(2t)+ B sin(2t))= e^{4t}((A+ B)cos(2t)+ (3B/2- A)sin(2t))$

x(0)= A= 2 and y(0)= A+ B= 2+ B= -3 so B= -5.

$x(t)= e^{4t}(-2 cos(2t)- 9 sin(2t))$
$y(t)= e^{4t}(-3 cos(2t)-(11/2)sin(2t)$.

 

[karush]

Once again, I would not use matrices for this simple a problem.

We have x'= 2x+ 2y so x''= 2x'+ 2y'.
We also have y'= -4x+ 6y so x''= 2x'+ 2(-4x+ 6y)= 2x'- 8x+ 12y.From x'= 2x+ 2y, 2y= x'- 2x so 12y= 6x'- 12x

We have x''= 2x'- 8x+ 6x'- 12x= 8x'- 20x
.x''- 8x'+ 20x= 0.

The characteristic equation is r^2- 8r+ 20= r^2- 8r+ 16- 16+ 20= (r- 4)^2+ 4= 0.
(r- 4)^2= -4 so $r- 4= \pm 2i$. $r= 4\pm 2i$

$x= e^{4t}(A cos(2t)+ B sin(2t))$
$12y= 6x'- 12x$ so $y= (1/2)x'- x$

$x'= 4e^{4t}(A cos(2t)+ B sin(2t))+ e^{4t}(-2A sin(2t)+ 2B cos(2t))= e^{4t}((4A+ 2B)cos(2t)+ (B- 2A)sin(2t))$

so $y= e^{4t}((2A+B)cos(2t)+ (B/2- A)sin(2t))- e^{4t}(A cos(2t)+ B sin(2t))= e^{4t}((A+ B)cos(2t)+ (3B/2- A)sin(2t))$

x(0)= A= 2 and y(0)= A+ B= 2+ B= -3 so B= -5.

$x(t)= e^{4t}(-2 cos(2t)- 9 sin(2t))$
$y(t)= e^{4t}(-3 cos(2t)-(11/2)sin(2t)$.

ok I don't think I understood every step, but let me try the the next one with this example on a new thread
it should be 3.4.5.7

that's a lot of help you are providing, I see too there are lots of views