# Use the method of "undetermined coefficients" to solve the 2nd ODE y''+4y'+4y=e^(-2x)sin2x ?

Dec 28, 2017

$y \left(x\right) = A x {e}^{- 2 x} + B {e}^{- 2 x} - \frac{1}{4} {e}^{- 2 x} \sin 2 x$

#### Explanation:

We have:

$y ' ' + 4 y ' + 4 y = {e}^{- 2 x} \sin 2 x$

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

$y ' ' + 4 y ' + 4 y$

And it's associated Auxiliary equation is:

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

Which has a repeated real root $m = - 2$

Thus the solution of the homogeneous equation is:

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

Particular Solution

With this particular equation [A], a probable solution is of the form:

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

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:

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

And:

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

 \ \ \ \ \ = e^(-2x)(-8b)cos2x + (8a)sin2x)

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

 {e^(-2x)(-8b)cos2x + (8a)sin2x)} + 4{e^(-2x)((2b-2a)cos2x - (2a+2b)sin2x)} + 4{e^(-2x)(acos2x + bsin2x)} = e^(-2x)sin2x

Equating coefficients of $\cos 2 x$ and $\sin 2 x$ we get:

$\cos 2 x : - 8 b + 8 b - 8 a + 4 a = 0$
$\sin 2 x : 8 a - 8 a - 8 b + 4 b = 1$

Solving simultaneous we have:

$a = 0 , b = - \frac{1}{4}$

And so we form the Particular solution:

${y}_{p} = - \frac{1}{4} {e}^{- 2 x} \sin 2 x$

General Solution

Which then leads to the GS of [A}

$y \left(x\right) = {y}_{c} + {y}_{p}$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \left(A x + B\right) {e}^{- 2 x} - \frac{1}{4} {e}^{- 2 x} \sin 2 x$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = A x {e}^{- 2 x} + B {e}^{- 2 x} - \frac{1}{4} {e}^{- 2 x} \sin 2 x$