Rewriting Higher Order Differential Equations in Normal Form

In many instances, higher order differential equations (i.e. of order 2 or higher) can be solved by reducing the higher order equation to a linear system (e.g. in dy/dt, or dx/dt or y' or x' )  in normal form. Recall that normal form yields a system such that:

dx1/dt = a11(t)x1 + a12 (t)x2 + ..a1n(t)xn + f1(t)
:
dx2/dt = a21 (t)x1 + a22 (t)x2 + ..a2n (t)xn + f2(t)
:
dx n/dt = an1 (t)x1 + an2 (t)x2 + ..ann(t) xn + fn(t)


Example: Rewrite the third order differential equation:

2y"'  - 6y"  + 4y' + y = sin t

in normal form.

Procedure:

First, divide thorugh by 2 and make y"' the subject to obtain:

y"' = - y/2  - 2y'  + 3y"  + sint / 2

Then use substitutions such that:

y = x1,  y' = x2, y" = x3

Now, since:

x1' = y' = x2

x2' = y" = x3

x3' = y"'

Then:

x1' = x2

x2' = x3

x3' = - x1/ 2 - 2x2 + 3x3  + 1/2 sin t


Problems for Math Mavens:

Rewrite these higher order DEs as systems in normal form:

1) y"  - 3y' + 4y = sin 3t


2)   2 d²y/ dt² + 4 dy/dt – 5y = 0