What is the complementary equation for this diffeq

[vande060]

Homework Statement

y" + 6y' + 9y = 1+x

Homework Equations

The Attempt at a Solution

r^2 + 6r + 9 = 0

(r +3)^2 = 0

r = -3

I thought the y_c = c₁e^-3x + c₂e^-3x

buy my prof says y_c = c₁e^-3x + c₂xe^-3x

why is the x in there?

 

[Mark44]

Homework Statement

y" + 6y' + 9y = 1+x

Homework Equations

The Attempt at a Solution

r^2 + 6r + 9 = 0

(r +3)^2 = 0

r = -3

I thought the y_c = c₁e^-3x + c₂e^-3x

buy my prof says y_c = c₁e^-3x + c₂xe^-3x

why is the x in there?

I agree with your prof. For a 2nd order, homogeneous differential equation, the solution space is two dimensional, which means that the complementary solution is all linear combinations of two linearly independent functions. Your complementary solutions is the same as (c₁ + c₂)e^-3x = Ke^-3x.

The characteristic equation for your homogeneous DE is r² + 6r + 9 = 0, which has repeated roots of r = -3.

The usual trick to get two linearly independent functions when the roots are repeated is to tack a factor of x onto the function. This gives {e^-3x, xe^-3x} as your set of linearly independent function.

This idea can be extended to higher order DEs. For example, if the DE is y''' + 6y'' + 12y' + 8y = 0, the characteristic equation is r³ + 6r² + 12r + 8 = 0, and this can be factored to (r + 2)³ = 0. Here the root r = -2 occurs three times.

A set of linearly independent functions is {e^-2x, xe^-2x, x²e^-2x}.

 

[vande060]

so whenever I have repeated roots I just have to add on the factor of x?

 

[Mark44]

Yes.

 

[vande060]

Yes.

you saved me again, thanks

 

[LCKurtz]

And it is easy to see why this works. If you call your differential equation L(y)=0 then, as you know

L(e^rx)=p(r)e^rx, which just says when you substitute e^rx into the DE, you get the characteristic polynomila p(r) times e^rx.

Look what happens if you differentiate this equation with respect to r (not x!), using the fact that differentiation with respect to r and x commute:

L(xe^rx)= p'(r)e^rx+ p(r)xe^rx

If r is a double root of the characteristic polynomial, both p(r) and p'(r) are zero, giving 0 on the right side. This says xe^rx is a solution to the DE.