# What is the general solution of the differential equation # y''' + 4y'' = 0 #?

##### 1 Answer

# y = Ax+B + Ce^(-4x) #

#### Explanation:

We have:

# y''' + 4y'' = 0 # ..... [A]

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

**Complimentary Function**

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

# m^3 +4m^2 = 0 #

# m^2(m+4) = 0 #

Which has repeated solutions

The roots of the axillary 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=e^(alphax)# ,#y_2=e^(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^(0x) + Ce^(-4x) #

# \ \ = Ax+B + Ce^(-4x) #

Note this solution has