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b.7.1.8 IVP with system of eq

karush

Well-known member
Jan 31, 2012
2,886
ck for typos
boyce 7.1 exercises
boyce 7.1 answers
Boyce Book

(a) Transform the given system into a single equation of second order.
(b) Find $x_1$ and $x_2$ that also satisfy the given initial conditions.
(c) Sketch the graph of the solution in the $x+1x_2$-plane for $t > 0$.
$\begin{array}{rrr}
x_1'=3x_1-2x_2 & x_1(0)=3\\
x_2'=2x_1-2x_2 & x_2(0)=\dfrac{1}{2}
\end{array}$
ok this is not a hw assignment but I reviewing before taking the class
also not sure if desmos can plot the answer
if there appears to be a typo go to the links above
the book seemed a little sparce on a good example to work with so...there was an exaple on page 362 but I couldn't follow it
well one way is to first rewrite x' to $x'=Ax$ where
$A=\left[\begin{array}{rrr}
3&-2\\
2&-2
\end{array}\right]$
so far

book answer
(a)$\quad x_1''-x_1'-2x_1=0$
(b)$\quad x_1=\dfrac{11}{3}e^{2t}-\dfrac{2}{3}e^{-t},\quad x_2=\dfrac{11}{6}e^{2t}-\dfrac{4}{3}e^{-t}$
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
You are given that $x_1'= 3x_1- 2x_2$ so $x_1''= 3x_1'- 2x_2'$.
You are also given that $x_2'= 2x_1- 2x_2$ so $x_1''= 3x_1'- 2(2x_1- 2x_2)= 3x_1'- 4x_1+ 4x_2$.

From $x_1'= 3x_1- 2x_2$, $2x_2= 3x_1- x_1'$ so
$x_1''= 3x_1'- 4x_1+ 4(3x_1- x_1')= -x_1'- 8x_1$

$x_1''+ x_1'+8x_1= 0$.
 

karush

Well-known member
Jan 31, 2012
2,886
You are given that $x_1'= 3x_1- 2x_2$ so $x_1''= 3x_1'- 2x_2'$.
You are also given that $x_2'= 2x_1- 2x_2$ so $x_1''= 3x_1'- 2(2x_1- 2x_2)= 3x_1'- 4x_1+ 4x_2$.

From $x_1'= 3x_1- 2x_2$, $2x_2= 3x_1- x_1'$ so
$x_1''= 3x_1'- 4x_1+ 4(3x_1- x_1')= -x_1'- 8x_1$

$x_1''+ x_1'+8x_1= 0$.
ok but the book answer $x_1''-x_1'-2x_1=0$

so if we rewrite the eq with $x=e^{γt}$ we have

$\left(\left(e^{γt}\right)\right)''\:-\left(\left(e^{γt}\right)\right)'\:-2e^{γt}=0$

not sure what we do with IV
 

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
731
You are given that $x_1'= 3x_1- 2x_2$ so $x_1''= 3x_1'- 2x_2'$.
You are also given that $x_2'= 2x_1- 2x_2$ so $x_1''= 3x_1'- 2(2x_1- 2x_2)= 3x_1'- 4x_1+ 4x_2$.

From $x_1'= 3x_1- 2x_2$, $2x_2= 3x_1- x_1'$ so
x_1''+ x_1'+8x_1= 0$.
This is an error. Since $2x_2= 3x_1- x_1'$, $4x_2= 6x_1- 2x_1'$
 

karush

Well-known member
Jan 31, 2012
2,886
ok I see the substitution