Question

What is the general solution of the differential equation `y''''-3y'''+3y''-y' = 0`?

Answer 1

`y = (Ax^2+Bx+C)e^x + D`

Explanation:

We have:

`y''''-3y'''+3y''-y' = 0` ..... [A]

This is a Fourth order linear 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.

Complementary Function

The Auxiliary equation associated with the homogeneous equation of [A] is (as indicated)(:

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

The challenge with higher order Differential Equation is solving the associated higher order Auxiliary equation. We note that `m` is a factor, so:

`m(m^3 -3m^2+3m-1) = 0`

By inspection we see `m=1` is a solution of the cubic, this by the factor theorem `m-1` is a factor of the cubic, so we can perform algebraic long division to get:

`m(m-1)^3 = 0`

Which has three real repeated roots `m=1` and a unique root `m=0`

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 a degree less than 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 [A} is of the form:

`y = (Ax^2+Bx+C)e^(1x) + De(0x)`
`\ \ = (Ax^2+Bx+C)e^x + D`

Note this solution has `4` constants of integration and `4` linearly independent solutions, hence their superposition is the General Solution.