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February 27th, 2017, 05:52 AM  #1 
Newbie Joined: Dec 2015 From: England Posts: 28 Thanks: 0  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

February 27th, 2017, 06:28 AM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,500 Thanks: 757 
what method are you supposed to use to solve these?

February 27th, 2017, 09:18 AM  #3 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,500 Thanks: 757 
$y^{\prime \prime} + 2 y^\prime + 2 y = x$ 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 $y_h(x) = c_1 e^{(1+i)x} + c_2 e^{(1i)x}$ 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 $y(x) = y_h(x) + y_p(x) = c_1 e^{(1+i)x} + c_2 e^{(1i)x} + \dfrac{x  1}{2}$ (b) is similarly solved. 
February 27th, 2017, 09:37 AM  #4 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,500 Thanks: 757 
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: 18,053 Thanks: 1395 
$\dfrac{\text{d}^2y}{\text{d}x^2}  4\dfrac{\text{d}y}{\text{d}x} + 4y = e^{2x}$ 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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