## Monday, August 5, 2013

### Tricks for Solving Higher Order Differential Equations (2) – Auxiliary Equations

Perhaps the neatest, easiest and most direct way to solve higher order differential equations is via the use of auxiliary equations. The best approach is to illustrate with a few examples:

Example (1):

Solve:   d2y/ dt2 - 10 dy/dt –25y = 0

Which can also be expressed: y” – 10y’ – 25y = 0

The auxiliary equation would be: m2  - 10 m – 25 = 0

In other words, the auxiliary equation is just a recasting as a quadratic but using the same coefficients as in the higher order DE. In this case we can factor to obtain: (m – 5) (m – 5)

This yields a repeated root of m = 5 so the solution is:

y = c1 exp (5x) + c2 exp(5x)

Example (2):

Solve:   2d2y/ dt2 - 5 dy/dt –3y = 0

Rewrite as: 2y” – 5y’ – 3y = 0

(Note: rewriting in prime form always makes it easier to spot the auxiliary)

Which yields the auxiliary eqn.: 2 m2  - 5 m – 3 = 0

Factor to get: 2 m2  - 5 m – 3  = (2m + 1) (m – 3)

Yielding the roots:  m =  - ½   and m = 3

So the general solution is: y = c1 exp(-x/2) + c2 exp(3x)

Before we go on, let’s take a second to reiterate here that each of the above examples is one of a higher order LINEAR differential equation. Recall here that a linear differential equation is one for which each term of the equation is either of the first degree in all the dependent variables and their various derivatives or does not contain any of them.

Thus: 2y” - 5y’ – 3y is a linear 2nd order differential equation

But:  3(y’)2 + y = 0

Is a non-linear 1st order differential equation since although the derivative is only of 1st degree the whole derivative is raised to the 2nd power. A homogenous linear differential equation, meanwhile, always possesses the trivial solution y = 0.

More examples:

Find the general solution of: 2y”’ – 5y” – y’ + 6y = 0

This requires an algebraic process of synthetic division, or very astute ability to see the factors since the auxiliary  is:

2 m3  - 5 m2 – m + 6 = 0

Using synthetic division:

2…..-5…..-1….6]  (-1)
…. ..-2…...7….-6
(2)..(-7)…(6)…(0)

After synthetic division the quadratic to solve becomes:

m2 – 7m/ 2  + 3 = 0  =  (m -2) (m – 3/2)

The interested reader can check that this is so by multiplying the above by (m +1) and simplifying.  You should get at the very end:

m3  - 5 m2 / 2 – m/ 2 + 3  = 0

Which – when multiplied through by (2)  - yields:  2 m3  - 5 m2 – m + 6 = 0

Thus the factors for the cubic are then: (m+1) (m- 3/2) (m- 2)

Yielding the general solution:

y=  c1 exp (-x) + c2 exp (2x) + c3 exp (3x/ 2)

Problems for Math Mavens:  Find the general solutions to each of the following:

1) d4y/ dt4 –2d3y/ dt3  –7 d2y/ dt2  + 20  dy/dt  –12 y = 0

2) y”’ + 3y” – 4y = 0