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y''+2y'+1=0 with initial conditions y(0)=1, y'(0)=-3

Mathematics
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Wolfram: \[\begin{array}{l} \text{Substitute }y(x)\text{ = }e^{\lambda x}\text{ into the differential equation:} \\ \frac{d^2\text{}}{dx^2}\left(e^{\lambda x}\right)+2 \frac{d\text{}}{dx}\left(e^{\lambda x}\right)\text{ = }0 \\ \text{Substitute }\frac{d^2\text{}}{dx^2}\left(e^{\lambda x}\right)\text{ = }\lambda ^2 e^{\lambda x}\text{ and }\frac{d\text{}}{dx}\left(e^{\lambda x}\right)\text{ = }\lambda e^{\lambda x}: \\ \lambda ^2 e^{\lambda x}+2 \lambda e^{\lambda x}\text{ = }0 \\ \text{Factor out }e^{\lambda x}: \\ \left(\lambda ^2+2 \lambda \right) e^{\lambda x}\text{ = }0 \\ \text{Since }e^{\lambda x}\neq 0\text{ for any finite }\lambda \text{, the zeros must come from the polynomial:} \\ \lambda ^2+2 \lambda \text{ = }0 \\ \text{Factor:} \\ \lambda (\lambda +2)\text{ = }0 \\ \text{Solve for }\lambda : \\ \lambda =-2\text{ or }\lambda =0 \\ \text{The \root }\lambda \text{ = }-2\text{ gives }y_1(x)\text{ = }c_1 e^{-2 x}\text{ as a solution, where }c_1\text{ is an arbitrary constant.} \\ \text{The \root }\lambda \text{ = }0\text{ gives }y_2(x)\text{ = }c_2\text{ as a solution, where }c_2\text{ is an arbitrary constant.} \\ \text{The general solution is the \sum of the \above solutions:} \\ y(x)\text{ = }y_1(x)+y_2(x)\text{ = }c_1 e^{-2 x}+c_2 \\ \text{Determine the particular solution \to }\frac{d^2y(x)}{dx^2}+2 \frac{dy(x)}{dx}=-1\text{ by the method of undetermined coefficients:} \\ \text{The particular solution \to }\frac{d^2y(x)}{dx^2}+2 \frac{dy(x)}{dx}\text{ = }-1\text{ is of the form:} \\ y_p(x)=a_1 x\text{, where }a_1\text{ was multiplied by }x\text{ \to account for }1\text{ \in the complementary solution.} \\ \text{Solve for the unknown constant} a_1: \\ \text{Compute }\frac{dy_p(x)}{dx}: \\ \begin{array}{c} \frac{dy_p(x)}{dx}\text{ = }\frac{d\text{}}{dx}\left(a_1 x\right) \\ \text{}\text{ = }a_1 \\\end{array} \\ \text{Compute }\frac{d^2y_p(x)}{dx^2}: \\ \begin{array}{c} \frac{d^2y_p(x)}{dx^2}\text{ = }\frac{d^2\text{}}{dx^2}\left(a_1 x\right) \\ \text{}\text{ = }0 \\\end{array} \\ \text{Substitute the particular solution }y_p(x)\text{ into the differential equation:} \\ \frac{d^2y_p(x)}{dx^2}+2 \frac{dy_p(x)}{dx}\text{ = }-1 \\ 2 a_1\text{ = }-1 \\ \text{Solve the equation:} \\ \begin{array}{l} a_1=-\frac{1}{2} \\\end{array} \\ \text{Substitute }a_1\text{ into }y_p(x)\text{ = }a_1 x: \\ y_p(x)\text{ = }-\frac{x}{2} \\ \text{The general solution is:} \\ y(x)\text{ = }y_{\text{c}}(x)+y_p(x)\text{ = }-\frac{x}{2}+c_1 e^{-2 x}+c_2 \\ \text{Solve for the unknown constants using the initial conditions:} \\ \text{Compute }\frac{dy(x)}{dx}: \\ \begin{array}{c} \frac{dy(x)}{dx}\text{ = }\frac{d\text{}}{dx}\left(-\frac{x}{2}+c_1 e^{-2 x}+c_2\right) \\ \text{}\text{ = }-2 c_1 e^{-2 x}-\frac{1}{2} \\\end{array} \\ \text{Substitute }y(0)=1\text{ into }y(x)\text{ = }-\frac{x}{2}+c_1 e^{-2 x}+c_2: \\ c_1+c_2=1 \\ \text{Substitute }y'(0)=-3\text{ into }\frac{dy(x)}{dx}\text{ = }-2 c_1 e^{-2 x}-\frac{1}{2}: \\ -2 c_1-\frac{1}{2}=-3 \\ \text{Solve the system:} \\ \begin{array}{c} \begin{array}{l} c_1=\frac{5}{4} \\ c_2=-\frac{1}{4} \\\end{array} \\\end{array} \\ \text{Substitute }c_1=\frac{5}{4}\text{ and }c_2=-\frac{1}{4}\text{ into }y(x)\text{ = }-\frac{x}{2}+c_1 e^{-2 x}+c_2: \\ y(x)=\frac{1}{4} \left(5 e^{-2 x}-2 x-1\right) \\\end{array}\]
this can't be right
hey that is y''+2y'+1=0 ? right? y''+2y'=-1 \[y=Ax^2+Bx+C\] \[y'=2Ax+B\] \[y''=2A\] Plug this in we get \[2A+2(2Ax+B)=-1\] WE need to find what A and B are ? 4Ax+2B+2A=-1 according to what this equation A and B should be what? Also when we are done with this we need to find the solution to y''+2y'=0

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Other answers:

can I just substitute y'' and y' with \[r^{2}\] and solve this quadratic equation \[r^{2}+2r+1=0\] where r=-1 and then \[y_{1}(t)=e^{-1t} y_{2}(t)=e^{-1t}\] so \[c_{1}e^{-1t}+c_{2}e^{-1t}=y(t)\] and then substitute initial values?
if you had this yes y''+2y'+y=0
but you have y''+2y'+1=0
oops you are right
so it's non-homogeneous I guess?
you could tackle it like y''+2y'=-1 with undetermined coefficients or something I bet

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