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February 27th, 2017, 06:52 AM   #1
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finding the general solution

I can't seem to find the general solution for part b as all the values of a, b or c just equal 0
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February 27th, 2017, 07:28 AM   #2
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what method are you supposed to use to solve these?
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February 27th, 2017, 10:18 AM   #3
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$y^{\prime \prime} + 2 y^\prime + 2 y = x$

first find homogeneous solution

$y^{\prime \prime} + 2 y^\prime + 2 y = 0$

the characteristic equation is

$s^2 + 2s +2 = 0$

with roots

$s = -1 \pm i$

so homogeneous solutions is given by

$y_h(x) = c_1 e^{(-1+i)x} + c_2 e^{(-1-i)x}$

using the method of undetermined cofficients we try

$y_p = c_3 x+c_4$

$y_p^\prime = c_3$

$y_p ^{\prime \prime} = 0$

$y_p ^{\prime \prime} + 2 y_p^\prime + 2 y_p = 0 + c_3 +2 c_3 x +2 c_4 =

2c_3 +2 c_4 +2 c_3 x = x$

so

$2c_3 +2 c_4 = 0$
$2c_3 = 1$

$c_3 = \dfrac 1 2$

$c_4 = -\dfrac 1 2$

and finally

$y(x) = y_h(x) + y_p(x) = c_1 e^{(-1+i)x} + c_2 e^{(-1-i)x} + \dfrac{x - 1}{2}$

(b) is similarly solved.
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February 27th, 2017, 10:37 AM   #4
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actually (b) is a bit messier but we use the same method

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

finding the homogeneous solution first we have characteristic polynomial

$s^2 - 4s + 4 = 0$

$s=2$ and this root is of order 2.

This results in a homogenous solution of

$y_h(x) = c_1 e^{2x} + c_2 x e^{2x}$

What about our particular solution? Ordinarily, given the driving function $e^{2x}$, we'd choose a form of

$y_p(x) = c_3 e^{2x}$

but this won't work as this is just a form of the homogeneous solution. Same with

$y_p(x) = c_3 e^{2x} + c_4 x e^{2x}$

so we have to try a solution of the form

$y_p(x) = c_3 x^2 e^{2x}$

$y_p^\prime(x) = 2 c_3 e^{2 x} x (x+1)$

$y_p^{\prime \prime} = 2 c_3 e^{2 x} \left(2 x^2+4 x+1\right)$

plugging all this back into the diff eq we get

$ 2 c_3 e^{2 x} \left(2 x^2+4 x+1\right) - 4\left(2 c_3 e^{2 x} x (x+1)\right) + 4c_3 x^2 e^{2x} = 2 c_3 e^{2 x} = e^{2x}$

so

$2 c_3 = 1$

$c_3 = \dfrac 1 2$

$y(x) = y_h(x)+y_p(x) = c_1 e^{2x} + c_2 x e^{2x} + \dfrac 1 2 x^2 e^{2x}$
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February 27th, 2017, 12:17 PM   #5
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$\dfrac{\text{d}^2y}{\text{d}x^2} - 4\dfrac{\text{d}y}{\text{d}x} + 4y = e^{2x}$ is equivalent to $e^{-2x}\dfrac{\text{d}^2y}{\text{d}x^2} - 4e^{-2x}\dfrac{\text{d}y}{\text{d}x} + 4e^{-2x}y = 1$.

Integrating that gives $e^{-2x}\dfrac{\text{d}y}{\text{d}x} - 2e^{-2x}y = x + \text{A}$, where $\text{A}$ is a constant.

Integrating again gives $e^{-2x}y = \frac12x^2 + \text{A}x + \text{B}$, where $\text{B}$ is a constant.

Hence the general solution is $y = \left(\frac12x^2 + \text{A}x + \text{B}\right)e^{2x}$.
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