# What is the general solution of the differential equation # y'''-y''+44y'-4=0 #?

##### 2 Answers

# y = A + e^(1/2x){Bcos(sqrt(15)/2x) + Csin(sqrt(15)/2x)} + x#

#### Explanation:

We have:

# y'''-y''+44y'-4=0 #

Or, Alternatively:

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

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

**Complementary Function**

The homogeneous equation associated with [A] is

# y'''-y''+4y' = 0 # ..... [B]

And it's associated Auxiliary equation is:

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

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

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

And we can complete the square:

# m((m-1/2)^2-(1/2)^2+4) = 0#

# :. m((m-1/2)^2+4-1/4) = 0#

# :. m((m-1/2)^2+15/4) = 0#

And so we have the solutions:

# m=0 #

# (m-1/2)^2 = -15/4 => m = 1/2+sqrt(15)/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 [B] is:

# y = Ae^(0x) + e^(1/2x){Bcos(sqrt(15)/2x) + Csin(sqrt(15)/2x)}#

# \ \ = A + e^(1/2x){Bcos(sqrt(15)/2x) + Csin(sqrt(15)/2x)}#

**Particular Solution**

In order to find a particular solution of the non-homogeneous equation:

# y'''-y''+4y' = f(x) \ \ # with#f(x)=4 # ..... [C]

then as

However, such a solution already exists in the CF solution and so must consider a potential solution of the form

Differentiating

# y' = a #

# y'' = 0 #

# y''' = 0 #

Substituting these results into the DE [A] we get:

# 0-0+4a = 4 => a=1 #

And so we form the Particular solution:

# y_p = x #

**General Solution**

Which then leads to the GS of [A}

# y(x) = y_c + y_p #

# \ \ \ \ \ \ \ = A + e^(1/2x){Bcos(sqrt(15)/2x) + Csin(sqrt(15)/2x)} + x#

Note this solution has