Question

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

Answer 1

`"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)`

Answer 2

`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, `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