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.




        y'' - 12y' + 36y  =  0



The characteristic equation is 

        r2 - 12r + 36  =  0


        (r - 6)2  =  0

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

        y1  =  e6t 

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)e6t 

and solve for v(t).  The derivatives are 

        y2'  =  v'(t)e6t + 6v(t)e6t  =  (v'(t) + 6v(t))e6t 

        y2''  =  (v''(t) + 6v'(t))e6t + 6(v'(t) + 6v(t))e6t =  (v''(t) + 12v'(t) + 36v(t))e6t 

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

        (v''(t) + 12v'(t) + 36v(t))e6t - 12[(v'(t) + 6v(t))e6t] + 36v(t)e6t 

         =  [v''(t) + 12v'(t) + 36v(t) - 12v'(t) - 72v(t) + 36v(t)]e6t 

         =  v''(t)e6t  =  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)e6t  =  te6t  

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

        W  =  (e6t)(e6t + 6te6t) - (6e6t)(te6t)  =  e12t + 6te12t 6te12t  =  e12t 

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

        y  =  c1e6t + c2te6t 


Theorem  (Solution for Repeated Roots)


            ay'' + by' + cy  =  0

be a differential equation such that the characteristic equation has the repeated root r.  That is

           b2 - 4ac  =  0

Then the general solution to the differential equation is given by 

          y  =  c1ert + c2tert 


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 among other places.



Find the solution to 

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



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.



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

        t2y'' + 2ty' - 2y  =  0

Find the general 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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