Repeated Roots and Reduction of Order

Now that we know how to solve second order linear homogeneous differential equations with constant coefficients such that the characteristic equation has distinct roots (either real or complex), the next task will be to deal with those which have repeated roots.  We proceed with an example.

Example

Solve

        y'' - 12y' + 36y  =  0

Solution

The characteristic equation is 

        r² - 12r + 36  =  0

or   

        (r - 6)²  =  0

We have only the root r  =  6 which gives the solution

        y₁  =  e^6t 

By general theory, there must be two linearly independent solutions to the differential equation.  We have found one and now search for a second.  Fortunately, a long time ago a mathematician named D'Alembert came up with a way to find the second linearly independent solution.  His idea was to write the second solution in the form

        y₂  =  v(t)e^6t 

and solve for v(t).  The derivatives are 

        y₂'  =  v'(t)e^6t + 6v(t)e^6t  =  (v'(t) + 6v(t))e^6t 

        y₂''  =  (v''(t) + 6v'(t))e^6t + 6(v'(t) + 6v(t))e^6t =  (v''(t) + 12v'(t) + 36v(t))e^6t 

Since y₂ is a solution to the differential equation, we can plug it in to get

        (v''(t) + 12v'(t) + 36v(t))e^6t - 12[(v'(t) + 6v(t))e^6t] + 36v(t)e^6t 

         =  [v''(t) + 12v'(t) + 36v(t) - 12v'(t) - 72v(t) + 36v(t)]e^6t 

         =  v''(t)e^6t  =  0

Since an exponential is never zero, we can conclude that 

        v''(t)  =  0

Integrating twice gives

        v(t)  =  c₂t + c₃ 

For c₂  =  0  and c₃  =  1, we get the original solution from the characteristic equation.  However, for c₂  =  1  and c₃  =  0, we get

        y₂  =  v(t)e^6t  =  te^6t  

We need to check that y₁ and y₂ are linearly independent.  We compute the Wronskian

        W  =  (e^6t)(e^6t + 6te^6t) - (6e^6t)(te^6t)  =  e^12t + 6te^12t 6te^12t  =  e^12t 

Since the Wronskian is an exponential, it is nonzero.  Hence the two solutions are linearly independent.  The general solution is 

        y  =  c₁e^6t + c₂te^6t 

To prove this theorem, we just go through the same steps as in the example above.  We will leave out the general proof here.  The proof can be found at http://www.math.colostate.edu/m340/notes/ODEI/node4.html among other places.

Example

Find the solution to 

        y'' + 10y' + 25y  =  0        y(0)  =  2,    y'(0)  =  3

Solution

The characteristic equation is 

        r² + 10r + 25  =  (r + 5)² =  0

This has the repeated root of r  =  -5.  The general solution is 

        y  =  c₁e^-5t + c₂te^-5t 

We use the initial values to determine the constants

        2  =  c₁e^-5(0) + c₂(0)e^-5(0)  =  c₁        

Taking the derivative gives

        y'  =  -10e^-5t + c₂(e^-5t - 5te^-5t)  

Plugging in the initial value gives

        3  =  -10 + c₂(1 - 0)  

        c₂  =  13

The final solution is 

        y  =  2e^-5t + 13te^-5t 

The graph is pictured below

Reduction of Order

When there are repeated roots, one of the linearly independent solutions was easy to find, while for the other solution we assumed that it had the form of a function times the known solution.  This approach works more generally.  

If y₁ is a known solution to a homogeneous linear differential equation, then we can seek a second linearly independent solution by writing 

        y₂  =  v(t)y₁

We demonstrate this with an example.

Example

Given that  y₁  =  t is a solution to the differential equation

        t²y'' + 2ty' - 2y  =  0

Find the general solution.

Solution

We write

        y₂  =  vt    y₂'  =  v't + v    y₂''  =  v''t + v' + v'  =  v''t + 2v'

Substituting back into the differential equation gives

        t²(v''t + 2v') + 2t(v't + v) - 2(vt)  =  0

        t³v'' + 2t²v' + 2t²v' + 2tv - 2tv  =  0

        t³v'' + 4t²v'  =  0

        tv'' + 4v'  =  0

Notice that the "v" term dropped out.  This will always be the case.  We can now let

        u  =  v'    u'  =  v''

Substituting gives

        tu' + 4u  =  0

This is a first order differential equation (hence the title "Reduction of Order").  We can separate to get

        du/u  =  -4dt/t

Now integrate to get

        ln|u|  =  -4ln|t|  + C

        u  =  Ct ^-4 

        v'  =  Ct ^-4 

        v  =  Ct ^-5

We can conclude that a second linearly independent solution is given by 

        y₂  =  (t^-3)(t)  =  t ^-2 

The general solution is 

        y  =  c₁t + c₂t ^-2 

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