anonymous
  • anonymous
y''+2y'+1=0 with initial conditions y(0)=1, y'(0)=-3
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
.Sam.
  • .Sam.
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}\]
anonymous
  • anonymous
this can't be right
myininaya
  • myininaya
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
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?
myininaya
  • myininaya
if you had this yes y''+2y'+y=0
myininaya
  • myininaya
but you have y''+2y'+1=0
anonymous
  • anonymous
oops you are right
anonymous
  • anonymous
so it's non-homogeneous I guess?
anonymous
  • anonymous
you could tackle it like y''+2y'=-1 with undetermined coefficients or something I bet

Looking for something else?

Not the answer you are looking for? Search for more explanations.