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

Feb 3, 2018

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

Feb 3, 2018

$y = A + {e}^{\frac{1}{2} x} \left\{B \cos \left(\frac{\sqrt{15}}{2} x\right) + C \sin \left(\frac{\sqrt{15}}{2} x\right)\right\} + x$

#### Explanation:

We have:

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

Or, Alternatively:

$y ' ' ' - y ' ' + 4 y ' = 4$ ..... [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

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

And it's associated Auxiliary equation is:

${m}^{3} - {m}^{2} + 4 m = 0$

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::

$m \left({m}^{2} - m + 4\right) = 0$

And we can complete the square:

$m \left({\left(m - \frac{1}{2}\right)}^{2} - {\left(\frac{1}{2}\right)}^{2} + 4\right) = 0$
$\therefore m \left({\left(m - \frac{1}{2}\right)}^{2} + 4 - \frac{1}{4}\right) = 0$
$\therefore m \left({\left(m - \frac{1}{2}\right)}^{2} + \frac{15}{4}\right) = 0$

And so we have the solutions:

$m = 0$
${\left(m - \frac{1}{2}\right)}^{2} = - \frac{15}{4} \implies m = \frac{1}{2} + \frac{\sqrt{15}}{2} -$

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 $m = \alpha , \beta , \ldots$ will yield linearly independent solutions of the form ${y}_{1} = A {e}^{\alpha x}$, ${y}_{2} = B {e}^{\beta x}$, ...
• Real repeated roots $m = \alpha$, will yield a solution of the form $y = \left(A x + B\right) {e}^{\alpha x}$ where the polynomial has the same degree as the repeat.
• Complex roots (which must occur as conjugate pairs) $m = p \pm q i$ will yield a pairs linearly independent solutions of the form $y = {e}^{p x} \left(A \cos \left(q x\right) + B \sin \left(q x\right)\right)$

Thus the solution of the homogeneous equation [B] is:

$y = A {e}^{0 x} + {e}^{\frac{1}{2} x} \left\{B \cos \left(\frac{\sqrt{15}}{2} x\right) + C \sin \left(\frac{\sqrt{15}}{2} x\right)\right\}$
$\setminus \setminus = A + {e}^{\frac{1}{2} x} \left\{B \cos \left(\frac{\sqrt{15}}{2} x\right) + C \sin \left(\frac{\sqrt{15}}{2} x\right)\right\}$

Particular Solution

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

$y ' ' ' - y ' ' + 4 y ' = f \left(x\right) \setminus \setminus$ with $f \left(x\right) = 4$ ..... [C]

then as $f \left(x\right)$ 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 $y = a x$, Where the constants $a$ is to be determined by direct substitution and comparison:

Differentiating $y = a x$ wrt $x$ we get:

$y ' = a$
$y ' ' = 0$
$y ' ' ' = 0$

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

$0 - 0 + 4 a = 4 \implies a = 1$

And so we form the Particular solution:

${y}_{p} = x$

General Solution

Which then leads to the GS of [A}

$y \left(x\right) = {y}_{c} + {y}_{p}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = A + {e}^{\frac{1}{2} x} \left\{B \cos \left(\frac{\sqrt{15}}{2} x\right) + C \sin \left(\frac{\sqrt{15}}{2} x\right)\right\} + x$

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