# Use the method of "undetermined coefficients" to solve the 2nd ODE y''-2y'=12e^(2x)-8e^(-2x)?

Dec 29, 2017

$y \left(x\right) = A + B {e}^{2 x} + 6 x {e}^{2 x} - {e}^{- 2 x}$

#### Explanation:

We have:

$y ' ' - 2 y ' = 12 {e}^{2 x} - 8 {e}^{- 2 x}$

This is a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, ${y}_{c}$ of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic 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

$y ' ' - 2 y ' = 0$

And it's associated Auxiliary equation is:

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

Which has distinct real root $m = 0$ and $m = 2$

Thus the solution of the homogeneous equation is:

${y}_{c} = A {e}^{0 x} + B {e}^{2 x}$
$\setminus \setminus \setminus = A + B {e}^{2 x}$

Particular Solution

With this particular equation [A], dur to the fact that ${e}^{2 x}$ is already part of the CF then a probable solution is of the form:

$y = a x {e}^{2 x} + b {e}^{- 2 x}$

Where $a , b$ are constant coefficients to be determined. Let us assume the above solution works, in which case be differentiating wrt $x$ we have:

$y ' \setminus \setminus = 2 a x {e}^{2 x} + a {e}^{2 x} - 2 b {e}^{- 2 x}$
$\setminus \setminus \setminus \setminus \setminus = \left(2 a x + a\right) {e}^{2 x} - 2 b {e}^{- 2 x}$

$y ' ' = 2 \left(2 a x + a\right) {e}^{2 x} + \left(2 a\right) {e}^{2 x} + 4 b {e}^{- 2 x}$
$\setminus \setminus \setminus \setminus \setminus = \left(4 a x + 4 a\right) {e}^{2 x} + 4 b {e}^{- 2 x}$

Substituting into the initial Differential Equation $\left[A\right]$ we get:

$\left\{\left(4 a x + 4 a\right) {e}^{2 x} + 4 b {e}^{- 2 x}\right\} - 2 \left\{\left(2 a x + a\right) {e}^{2 x} - 2 b {e}^{- 2 x}\right\} = 12 {e}^{2 x} - 8 {e}^{- 2 x}$

$\therefore 4 a x {e}^{2 x} + 4 a {e}^{2 x} + 4 b {e}^{- 2 x} - 4 a x {e}^{2 x} - 2 a {e}^{2 x} + 4 b {e}^{- 2 x} = 12 {e}^{2 x} - 8 {e}^{- 2 x}$

$\therefore 4 a {e}^{2 x} + 4 b {e}^{- 2 x} - 2 a {e}^{2 x} + 4 b {e}^{- 2 x} = 12 {e}^{2 x} - 8 {e}^{- 2 x}$

Equating coefficients of ${e}^{2 x}$ and ${e}^{- 2 x}$ we get:

${e}^{2 x} \setminus \setminus \setminus : 4 a - 2 a = 12 \implies a = 6$
${e}^{- 2 x} : 4 b + 4 b = - 8 \implies b = - 1$

And so we form the Particular solution:

${y}_{p} = 6 x {e}^{2 x} - {e}^{- 2 x}$

General Solution

Which then leads to the GS of [A}

$y \left(x\right) = {y}_{c} + {y}_{p}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = A + B {e}^{2 x} + 6 x {e}^{2 x} - {e}^{- 2 x}$