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
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