# What is the general solution of the differential equation # y''' - 3y'-2y =0 #?

##### 2 Answers

#y = c_1e^(-x) + c_2xe^(-x) + c_3e^(2x)#

Well, this is a **linear third-order ordinary differential equation**. We can assume a solution of

#y = e^(rx)#

so that

#r^3 e^(rx) - 3re^(rx) - 2e^(rx) = 0#

And since

#r^3 - 3r - 2 = 0#

Now we need to figure out how this can be factored. By synthetic division, we have a guess for

#ul(-1)|" "1" "" "0" "-3" "-2#

#+" "" "" "" "-1" "" "1" "" "2#

#"----------------------------------------"#

#" "" "" "1" "-1" "-2" "" "0#

And we happen to be correct. Thus, we obtain a factorization of:

#(r+1)(r^2 - r - 2) = 0#

#(r+1)^2(r-2) = 0#

So, the roots are **linear combination** of these three roots (i.e. not all the coefficients

#color(blue)(y = c_1e^(-x) + c_2xe^(-x) + c_3e^(2x))# Note that the multiplicity-2 root has two unique constants in its part of the linear combination, but the second root instance is distinguished by another

#x# term in front.

And we should check whether we are correct or not... Do we still have that

#y''' - 3y' - 2y = 0# ?

Firstly, we have a straightforward differentiation using the product rule for the first term:

#y''' = -c_1e^(-x) + c_2(d^2)/(dx^2)[-xe^(-x) + e^(-x)] + 8c_3e^(2x)#

#= -c_1e^(-x) + c_2(d)/(dx)[-[-xe^(-x) + e^(-x)] - e^(-x)] + 8c_3e^(2x)#

#= -c_1e^(-x) + c_2(d)/(dx)[xe^(-x) - 2e^(-x)] + 8c_3e^(2x)#

#= -c_1e^(-x) + c_2[[-xe^(-x) + e^(-x)] + 2e^(-x)] + 8c_3e^(2x)#

#= -c_1e^(-x) - c_2[xe^(-x) - 3e^(-x)] + 8c_3e^(2x)#

And for the second term:

#y' = -c_1e^(-x) - c_2[xe^(-x) - e^(-x)] + 2c_3e^(2x)#

So, we shall verify whether our solution works.

#y''' - 3y' - 2y stackrel(?" ")(=) 0#

#= {-c_1e^(-x) - c_2[xe^(-x) - 3e^(-x)] + 8c_3e^(2x)} - 3{-c_1e^(-x) - c_2[xe^(-x) - e^(-x)] + 2c_3e^(2x)} - 2{c_1e^(-x) + c_2xe^(-x) + c_3e^(2x)}#

#= -c_1e^(-x) + c_2xe^(-x) - 3c_2e^(-x) + 8c_3e^(2x) + 3c_1e^(-x) + 3c_2xe^(-x) - 3c_2e^(-x) - 6c_3e^(2x) - 2c_1e^(-x) - 2c_2xe^(-x) - 2c_3e^(2x)#

#= -c_1e^(-x) - c_2xe^(-x) + cancel(3c_2e^(-x)) + cancel(8c_3e^(2x)) + 3c_1e^(-x) + 3c_2xe^(-x) - cancel(3c_2e^(-x)) - cancel(6c_3e^(2x)) - 2c_1e^(-x) - 2c_2xe^(-x) - cancel(2c_3e^(2x))#

#= -c_1e^(-x) - cancel(c_2xe^(-x)) + 3c_1e^(-x) + cancel(3c_2xe^(-x)) - 2c_1e^(-x) - cancel(2c_2xe^(-x))#

#= -cancel(c_1e^(-x)) + cancel(3c_1e^(-x)) - cancel(2c_1e^(-x))#

#= 0# #color(blue)(sqrt"")#

So indeed, we **do** have the general solution.

# y = Axe^(-x) + Be^(-x) + Ce^(2x)#

#### Explanation:

We have:

# y''' - 3y'-2y =0 # ..... [A]

This is a **Third** order **linear** Homogeneous Differentiation Equation with constant coefficients. The standard approach is to find a solution,

**Complementary Function**

The Auxiliary equation associated with the homogeneous equation of [A] is:

# m^3 -3m-2 = 0 #

The challenge with higher order Differential Equation is solving the associated higher order Auxiliary equation. By inspection we see

# (m+1)(m^2-m-2) = 0#

# :. (m+1)(m+1)(m-2) = 0#

# :. (m+1)^2(m-2) = 0#

Which has three real solutions, one distinct and one repeated twice:

#m=-1,-1,2# .

The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.

- Real distinct roots
#m=alpha,beta, ...# will yield linearly independent solutions of the form#y_1=Ae^(alphax)# ,#y_2=Be^(betax)# , ... - Real repeated roots
#m=alpha# , will yield a solution of the form#y=(Ax+B)e^(alphax)# where the polynomial has the same degree as the repeat. - Complex roots (which must occur as conjugate pairs)
#m=p+-qi# will yield a pairs linearly independent solutions of the form#y=e^(px)(Acos(qx)+Bsin(qx))#

Thus the solution of the homogeneous equation is:

# y = (Ax+B)e^(-1x) + Ce^(2x)#

# \ \ = Axe^(-x) + Be^(-x) + Ce^(2x)#

Note this solution has