ODE4.1

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$y''+2y'+5y=e^{-x}\sin(2x)\,$

First, solve the homogeneous equation to get $y h$ . Then, guess a particular solution $y p$ and add them together to get the general solution $y = y h + y p$ .

$y_h''+2y_h'+5y_h=0\,$

Set up the characteristic equation and solve.

$m^2+2m+5=0\,$

So,

$m = \frac{-2 \pm \sqrt{4-20} }{2}\,$

$m = -1 \pm 2i\,$

Therefore, the solution to the homogeneous equation is

$y_h=c_1 e^{-x} \cos(2x) + c_2 e^{-x} \sin(2x)\,$
$y_h=e^{-x}[c_1 \cos(2x) + c_2 \sin(2x)]\,$

Now, check the homogeneous solution to see if any new information can be gained. Start by taking the first and second derivatives of $y h$ .

$y_h' = e^{-x}[ (-2c_1-c_2)\sin(2x) + (-c_1+2c_2)\cos(2x) ] \,$

$y_h''= e^{-x}[ (-4c_1-2c_2)\cos(2x)-(-2c_1+4c_2)\sin(2x) ] - e^{-x}[ (-2c_1-c_2)\sin(2x) + (-c_1+2c_2)\cos(2x) ]\,$
$y_h''= e^{-x}[ (-3c_1-4c_2)\cos(2x) + (4c_1-3c_2)\sin(2x) ]\,$

Substitution of these derivatives into the homogeneous version of the original differential equation and collection of the exponential and trig functions gives the following relations.

$e^{-x}\cos(2x)[-4c_1+c_1-4c_2-2c_1+2c_2+5c_1]=0\,$
$e^{-x}\sin(2x)[4c_1-3c_2-4c_1-2c_2+5c_2]=0\,$

So from the first equation, $c 1$ is arbitrary and $- 2 c 2 = 0$ , so $c 2 = 0$ . From the second equation, no new information is obtained. Therefore, a better homogeneous equation is

$y_h = c_1 e^{-x} \cos(2x)\,$

Now a particular solution should be guessed. Based on the nonhomogeneous part of the original DE, $e - x sin(2 x)$ , a first guess is

$y_p=Ae^{-x}[c_3 \cos(2x) + c_4 \sin(2x)]\,$ , with $A$ , $c 3$ , and $c 4$ undetermined constants.

But, the homogeneous solution appears as one of the terms of the guessed particular solution, so that term should be multiplied by $x$ . Now, a guess for the particular solution is

$y_p = A e^{-x} [ c_3 x \cos(2x) + c_4 \sin(2x) ]\,$

Find the first and second derivatives of $y p$ .

$y_p'= A e^{-x} [ -2 c_3 x \sin(2x) + c_3 \cos(2x) + 2 c_4 \cos(2x) ] - A e^{-x} [ c_3 x \cos(2x) + c_4 \sin(2x) ]\,$
$y_p'= A e^{-x} [ -c_3 x \cos(2x) -2 c_3 x \sin(2x) + (c_3+2c_4) \cos(2x) - c_4 \sin(2x) ]\,$

$y_p''=A e^{-x} [ 2 c_3 x \sin(2x) -c_3 \cos(2x) -4 c_3 x \cos(2x) -2 c_3 \sin(2x) -2(c_3+2c_4)\sin(2x) -2c_4\cos(2x) ] \,$
$-A e^{-x}[ -c_3 x \cos(2x) -2 c_3 x \sin(2x) + (c_3+2c_4) \cos(2x) - c_4 \sin(2x) ]\,$
$y_p''=A e^{-x} [ -3 c_3 x \cos(2x) + 4 c_3 x \sin(2x) + (-2c_3 -4c_4)\cos(2x) + (-4c_3-3c_4)\sin(2x) ] = e^{-x} \sin(2x)\,$

Substitute these derivatives into the original DE to get these relations:

$A e^{-x} [ (-3c_3-2c_3+5c_3) x \cos(2x) + (2c_3+2c_3-4c_3) x \sin(2x) + \,$
$(-c_3-2c_4-c_3-2c_4+2c_3+4c_4) \cos(2x) + (-4c_3-4c_4+c_4-2c_4+5c_4) \sin(2x) ]\,$

So comparing the coefficients of $x sin(2 x)$ and $x cos(2 x)$ gives no information about $c 3$ except that it is arbitrary so far. The other terms give the relations

$-4 c_3 = 1\,$ so that $c_3 = \frac{-1}{4}\,$ , and
$c 4$ is arbitrary, so that the particular solution is

$y_p = e^{-x} [ -\frac{1}{4} x \cos(2x) + c_4 \sin(2x) ]\,$

Finally, the general solution is

$y(x) = y_h + y_p = e^{-x} [ - \frac{1}{4} x \cos(2x) + c_1 \cos(2x) + c_4 \sin(2x) ] \,$

Ordinary Differential Equations

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ODE4.1

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