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 Differential Equations Ordinary and Partial Differential Equations Math Forum

February 27th, 2017, 05:52 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
Attached Images Capture2.PNG (8.2 KB, 11 views) February 27th, 2017, 06:28 AM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,582 Thanks: 1427 what method are you supposed to use to solve these? February 27th, 2017, 09:18 AM #3 Senior Member   Joined: Sep 2015 From: USA Posts: 2,582 Thanks: 1427 $y^{\prime \prime} + 2 y^\prime + 2 y = x$ 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, 09:37 AM #4 Senior Member   Joined: Sep 2015 From: USA Posts: 2,582 Thanks: 1427 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, 11:17 AM #5 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 $\dfrac{\text{d}^2y}{\text{d}x^2} - 4\dfrac{\text{d}y}{\text{d}x} + 4y = e^{2x}$ 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 Tags finding, general, solution Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Zynoakib Calculus 3 October 16th, 2013 02:01 PM Zynoakib Algebra 1 October 16th, 2013 01:59 PM manchester20 Economics 1 August 23rd, 2012 02:47 AM dunn Differential Equations 2 February 19th, 2012 03:38 AM manchester20 Applied Math 0 December 31st, 1969 04:00 PM

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