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

#"Characteristic equation is : "#
#z^3 - z^2 + 4 z = 0#
#=> z(z^2 - z + 4) = 0#
#=> z = 0 " OR " z^2 - z + 4 = 0#
#"disc of the quad. eq. = 1 - 16 = -15 < 0"#
#"so we have two complex solutions, they are"#
#z = (1 pm sqrt(15) i)/2#
#"So the general solution of the homogeneous equation is : "#
#A + B' exp(x/2) exp((sqrt(15)/2) i x) +#
#C' exp(x/2) exp(-(sqrt(15)/2) i x)#
#= A + B exp(x/2) cos(sqrt(15)x /2) + C exp(x/2) sin(sqrt(15) x/2)#
#"The particular solution to the complete equation is"#
#"y=x ,"#
#"That is easy to see."#
#"So the complete solution is :"#
#y(x) = x + A + B exp(x/2) cos(sqrt(15) x/2) + C exp(x/2) sin(sqrt(15) x/2)#

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, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the polynomial equation with the coefficients of the derivatives., and then finding an independent particular solution, #y_p# of the non-homogeneous equation.

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# is a factor, so we get::

# 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 #f(x)# is a polynomial of degree #0#, we would look for a polynomial solution of the same degree, i.e. of the form #y = a#

However, such a solution already exists in the CF solution and so must consider a potential solution of the form #y=ax#, Where the constants #a# is to be determined by direct substitution and comparison:

Differentiating #y=ax# wrt #x# we get:

# 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 #3# constants of integration and #3# linearly independent solutions, hence by the Existence and Uniqueness Theorem their superposition is the General Solution