# What is the general solution of the differential equation  y''' - 3y'-2y =0 ?

Jul 9, 2017

$y = {c}_{1} {e}^{- x} + {c}_{2} x {e}^{- x} + {c}_{3} {e}^{2 x}$

Well, this is a linear third-order ordinary differential equation. We can assume a solution of

$y = {e}^{r x}$

so that

${r}^{3} {e}^{r x} - 3 r {e}^{r x} - 2 {e}^{r x} = 0$

And since ${e}^{r x} \ne 0$, we obtain the auxiliary equation:

${r}^{3} - 3 r - 2 = 0$

Now we need to figure out how this can be factored. By synthetic division, we have a guess for $r + 1$ as one factor:

$\underline{- 1} | \text{ "1" "" "0" "-3" } - 2$
$+ \text{ "" "" "" "-1" "" "1" "" } 2$
$\text{----------------------------------------}$
$\text{ "" "" "1" "-1" "-2" "" } 0$

And we happen to be correct. Thus, we obtain a factorization of:

$\left(r + 1\right) \left({r}^{2} - r - 2\right) = 0$

${\left(r + 1\right)}^{2} \left(r - 2\right) = 0$

So, the roots are $r = - 1$ (multiplicity $2$) and $r = 2$. This gives a general solution as a linear combination of these three roots (i.e. not all the coefficients ${c}_{i}$ are zero).

$\textcolor{b l u e}{y = {c}_{1} {e}^{- x} + {c}_{2} x {e}^{- x} + {c}_{3} {e}^{2 x}}$

Note that the multiplicity-2 root has two unique constants in its part of the linear combination, but the second root instance is distinguished by another $x$ term in front.

And we should check whether we are correct or not... Do we still have that

$y ' ' ' - 3 y ' - 2 y = 0$?

Firstly, we have a straightforward differentiation using the product rule for the first term:

$y ' ' ' = - {c}_{1} {e}^{- x} + {c}_{2} \frac{{d}^{2}}{{\mathrm{dx}}^{2}} \left[- x {e}^{- x} + {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}$

$= - {c}_{1} {e}^{- x} + {c}_{2} \frac{d}{\mathrm{dx}} \left[- \left[- x {e}^{- x} + {e}^{- x}\right] - {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}$

$= - {c}_{1} {e}^{- x} + {c}_{2} \frac{d}{\mathrm{dx}} \left[x {e}^{- x} - 2 {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}$

$= - {c}_{1} {e}^{- x} + {c}_{2} \left[\left[- x {e}^{- x} + {e}^{- x}\right] + 2 {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}$

$= - {c}_{1} {e}^{- x} - {c}_{2} \left[x {e}^{- x} - 3 {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}$

And for the second term:

$y ' = - {c}_{1} {e}^{- x} - {c}_{2} \left[x {e}^{- x} - {e}^{- x}\right] + 2 {c}_{3} {e}^{2 x}$

So, we shall verify whether our solution works.

y''' - 3y' - 2y stackrel(?" ")(=) 0

$= \left\{- {c}_{1} {e}^{- x} - {c}_{2} \left[x {e}^{- x} - 3 {e}^{- x}\right] + 8 {c}_{3} {e}^{2 x}\right\} - 3 \left\{- {c}_{1} {e}^{- x} - {c}_{2} \left[x {e}^{- x} - {e}^{- x}\right] + 2 {c}_{3} {e}^{2 x}\right\} - 2 \left\{{c}_{1} {e}^{- x} + {c}_{2} x {e}^{- x} + {c}_{3} {e}^{2 x}\right\}$

$= - {c}_{1} {e}^{- x} + {c}_{2} x {e}^{- x} - 3 {c}_{2} {e}^{- x} + 8 {c}_{3} {e}^{2 x} + 3 {c}_{1} {e}^{- x} + 3 {c}_{2} x {e}^{- x} - 3 {c}_{2} {e}^{- x} - 6 {c}_{3} {e}^{2 x} - 2 {c}_{1} {e}^{- x} - 2 {c}_{2} x {e}^{- x} - 2 {c}_{3} {e}^{2 x}$

$= - {c}_{1} {e}^{- x} - {c}_{2} x {e}^{- x} + \cancel{3 {c}_{2} {e}^{- x}} + \cancel{8 {c}_{3} {e}^{2 x}} + 3 {c}_{1} {e}^{- x} + 3 {c}_{2} x {e}^{- x} - \cancel{3 {c}_{2} {e}^{- x}} - \cancel{6 {c}_{3} {e}^{2 x}} - 2 {c}_{1} {e}^{- x} - 2 {c}_{2} x {e}^{- x} - \cancel{2 {c}_{3} {e}^{2 x}}$

$= - {c}_{1} {e}^{- x} - \cancel{{c}_{2} x {e}^{- x}} + 3 {c}_{1} {e}^{- x} + \cancel{3 {c}_{2} x {e}^{- x}} - 2 {c}_{1} {e}^{- x} - \cancel{2 {c}_{2} x {e}^{- x}}$

$= - \cancel{{c}_{1} {e}^{- x}} + \cancel{3 {c}_{1} {e}^{- x}} - \cancel{2 {c}_{1} {e}^{- x}}$

$= 0$ color(blue)(sqrt"")

So indeed, we do have the general solution.

Jul 9, 2017

$y = A x {e}^{- x} + B {e}^{- x} + C {e}^{2 x}$

#### Explanation:

We have:

$y ' ' ' - 3 y ' - 2 y = 0$ ..... [A]

This is a Third 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:

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

The challenge with higher order Differential Equation is solving the associated higher order Auxiliary equation. By inspection we see $m = - 1$ is a solution, so we can factor out $\left(m + 1\right)$ and use polynomial division (or inspection) to get a quadratic factor:

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

Which has three real solutions, one distinct and one repeated twice:

$m = - 1 , - 1 , 2$.

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 is:

$y = \left(A x + B\right) {e}^{- 1 x} + C {e}^{2 x}$
$\setminus \setminus = A x {e}^{- x} + B {e}^{- x} + C {e}^{2 x}$

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