General Solutions of Linear Equations

General Solutions of Linear Equations

Example 1. Find the general solution to the ordinary differential equation

y⁽⁴⁾ − 16y = 0

Solution. The characteristic polynomial is

r ⁴ − 16

It can be factorized as

r ⁴ − 16 = (r² − 4)(r² + 4) = (r − 2)(r + 2)(r − 2i)(r + 2i).

So we have two real roots and two complex roots . Each
root has the multiplicity one. So the general solution is



Example 2. Find a general solution to equation

5y⁽⁴⁾ + 3y⁽³⁾ = 0.

Solution. The characteristic polynomial is



We have on root of multiplicity 3 and the root of multiplicity
one. The general solution is



Example 3.Find a general solution to equation

y⁽³⁾ + 27y = 0

Solution. The characteristic polynomial is

r³ + 27

It can be factorized as



The general solution is



Example 3. Solve the initial value problem

y'' − 4y' + 3y = 0, y(0) = 7, y'(0) = 11.

Solution. First we find a general solution to this equation. The characteristic
polynomial is

r² − 4r + 3 = (r − 1)(r − 3)

Hence the general solution is



The derivative of this function is



Using the initial conditions we obtain the system of linear equations



Hence and . So the solution to the initial value problem is



Example 4. Solve the initial value problem

y(3) + 10y'' + 25y', y(0) = 3, y'(0) = 4, y''(0) = 5

Solution. The characteristic equation is

r ³ + 10r² + 25r = r(r + 5)²

The root has the multiplicity one and the root has the
multiplicity two.

So the general solution is



The first and the second derivatives of this solution are



and



Using the initial conditions we obtain the system of linear equations



Hence and and . Hence the solution is