# [SOLVED]IVP

#### karush

##### Well-known member
Change the second-order IVP into a system of equations
$y''+y'-2y=0\quad y(0)= 2\quad y'(0)=0$
let $x_1=y$ and $x_2=y'$ then $x_1'= x_2$ and $y''=x_2'$
then by substitution
$x_2'+x_2-2x_1=0$
then the system of first order of equations
$x_1'=x_2$
$x_2'=-x_2+2x_1$

hopefully so far..

#### Country Boy

##### Well-known member
MHB Math Helper
Yes, that is correct.

Now, what are $x_1(0)$ and $x_2(0)$?

• karush

#### karush

##### Well-known member
Yes, that is correct.
Now, what are $x_1(0)$ and $x_2(0)$?
$x_1'=x_2=y(0)=0$
and
$x_2'=-x_2+2x_1=0+2(2)=4$

its like chasing a rabbit in the briers

#### Country Boy

##### Well-known member
MHB Math Helper
Frankly, I am not sure what you are doing, You were told that y(0)= 2 and y'(0)= 0.

Since you defined $x_1(t)$ to be y(t) and $x_2(t)$ to be y'(t),
$x_1(0)= y(0)= 2$ and $x_2(0)= y'(0)= 0$

• karush

#### karush

##### Well-known member
Frankly, I am not sure what you are doing, You were told that y(0)= 2 and y'(0)= 0.

Since you defined $x_1(t)$ to be y(t) and $x_2(t)$ to be y'(t),
$x_1(0)= y(0)= 2$ and $x_2(0)= y'(0)= 0$
i think I get confused looking at multiple examples with all these different substitutions