Higher Order Homogeneous Differential Equations With Constant Coefficients

We now investigate how to solve higher order homogeneous linear differential equations with constant coefficients.  The good news is that we use the same technique that we used for second order differential equations.  We demonstrate this by an example.

Example

Solve

        y''' - 4y'' + y' + 6y  =  0

Solution

We make the assumption that the solution has the form

       y  =  e^rt

Then

        y' =  re^rt         y''  =  r²e^rt         y'''  =  r³e^rt

Substituting back into the differential equation, we get

        r³e^rt - 4(r²e^rt) + re^rt + 6e^rt   =  0

Dividing by e^rt, we get the characteristic equation

        r³ - 4r² + r + 6  =  0

We can factor to find the roots of the equation.  A calculator can efficiently do this, or you can use the rational root theorem to get

        (r + 1)(r - 2)(r - 3)  =  0

       r  =  -1        r  =  2        or       r  =  3

The general solution is

       y  =  c₁e^-t + c₂e^2t + c₃e^3t

Example

Find the solution to

        y(^v) - 4y'''  =  0        y(0)  =  2    y'(0)  =  3    y''(0)  =  1    y'''(0)  =  -1,    y^(iv)(0)  =  1

Solution

We have the characteristic equation

        r⁵ + 4r³  =  0   

Which has a root of order 3 at

        r  =  0

and complex roots

       r  =  2i     and    r  =  -2i

We use what we have learned about repeated roots and complex roots to get the general solution.  Since the order of the repeated root is 3, we have

        y₁  =  1,        y₂  =  t,        y₃  =  t²    For order 2, multiply by t and t²

The complex roots give the other two solutions

        y₃  =  cos(2t)       and       y₅  =  sin(2t)

The general solution is

        y  =  c₁ + c₂t + c₃t² + c₄ cos(2t) + c₅ sin(2t)

Now Find the first four derivatives

        y'  =  c₂ + 2c₃t - 2c₄ sin(2t) + 2c₅ cos(2t)

        y''  =  2c₃ - 4c₄ cos(2t) - 4c₅ sin(2t)

        y'''  =  8c₄ sin(2t) - 8c₅ cos(2t)

        y^(iv)  =  16c₄ cos(2t) + 16c₅ sin(2t)

Now plug in the initial conditions to get

        2  =  c₁ + c₄

        3  =  c₂  + 2c₅

        1  =  2c₃ - 4c₄

        -1  =  8c₅

        1  =  16c₄

We can solve this either by hand or by using a calculator to get

        c₁  =   31/16        c₂  =  23/4          c₃  =  5/8          c₄   =  1/16       c₅  = 1/8

The solution is

        y   =   31/16  +  23/4 t  +  5/8 t²  +  1/16 cos(2t)  +  1/8 sin(2t)

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