# 'write the following as a system of higher-order equations in standard form'

• Apr 23rd 2010, 10:18 PM
figs
'write the following as a system of higher-order equations in standard form'
It's been a long time since I've taken differential equations, and in the mathematical programming course I'm in, we're just about to get into using MATLAB to solve ODEs. We were given some diffEQ problems to solve by hand to refresh our memories. This is the only one I haven't been able to do:

"Write the system of equations
$\frac{d^2x}{dt^2} + \frac{d^3y}{dt^3} + 2\frac{d^2y}{dt^2} + x = t$

$
\frac{d^2x}{dt^2} + 2\frac{d^3y}{dt^3} + \frac{dx}{dt} + y = t^2
$

as a system of two higher-order equations in standard form."

I have no idea what the question is asking. I remember "standard form" being where you let $x_{1} = x, x_{2} = \dot{x}$, etc, and express the highest-order terms as $x^{(n)} = \dot{x}_{n-1}$, but the purpose of that is to make it so there are no higher-order terms, so 'higher-order equations' cannot refer to that.

thanks.
• May 20th 2010, 12:37 AM
CaptainBlack
Quote:

Originally Posted by figs
It's been a long time since I've taken differential equations, and in the mathematical programming course I'm in, we're just about to get into using MATLAB to solve ODEs. We were given some diffEQ problems to solve by hand to refresh our memories. This is the only one I haven't been able to do:

"Write the system of equations
$\frac{d^2x}{dt^2} + \frac{d^3y}{dt^3} + 2\frac{d^2y}{dt^2} + x = t$

$
\frac{d^2x}{dt^2} + 2\frac{d^3y}{dt^3} + \frac{dx}{dt} + y = t^2
$

as a system of two higher-order equations in standard form."

I have no idea what the question is asking. I remember "standard form" being where you let $x_{1} = x, x_{2} = \dot{x}$, etc, and express the highest-order terms as $x^{(n)} = \dot{x}_{n-1}$, but the purpose of that is to make it so there are no higher-order terms, so 'higher-order equations' cannot refer to that.

thanks.

Here you should try to rewite these in the form:

$
D^3y=f(t,x,y,Dx,Dy,D^2y)
$

$
D^2x=g(t,x,y,Dx,Dy,D^2y)
$

or possibly as:

$
D^3y=f(t,x,y,Dx,Dy,D^2x,D^2y)
$

$
D^3x=g(t,x,y,Dx,Dy,D^2x,D^2y)
$

CB