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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 Senior Member     Joined: Sep 2015 From: USA Posts: 1,655 Thanks: 841 what method are you supposed to use to solve these?
 February 27th, 2017, 10:18 AM #3 Senior Member     Joined: Sep 2015 From: USA Posts: 1,655 Thanks: 841 $y^{\prime \prime} + 2 y^\prime + 2 y = x$ 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. Thanks from topsquark
 February 27th, 2017, 10:37 AM #4 Senior Member     Joined: Sep 2015 From: USA Posts: 1,655 Thanks: 841 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 $2 c_3 = 1$ $c_3 = \dfrac 1 2$ $y(x) = y_h(x)+y_p(x) = c_1 e^{2x} + c_2 x e^{2x} + \dfrac 1 2 x^2 e^{2x}$
 February 27th, 2017, 12:17 PM #5 Global Moderator   Joined: Dec 2006 Posts: 18,247 Thanks: 1439 $\dfrac{\text{d}^2y}{\text{d}x^2} - 4\dfrac{\text{d}y}{\text{d}x} + 4y = e^{2x}$ 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 $y = \left(\frac12x^2 + \text{A}x + \text{B}\right)e^{2x}$. Thanks from fysmat

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