Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

.Sam. Group TitleBest ResponseYou've already chosen the best response.0
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 x1\right) \\\end{array}\]
 2 years ago

Timoha Group TitleBest ResponseYou've already chosen the best response.0
this can't be right
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
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
 2 years ago

Timoha Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
if you had this yes y''+2y'+y=0
 2 years ago

myininaya Group TitleBest ResponseYou've already chosen the best response.3
but you have y''+2y'+1=0
 2 years ago

Timoha Group TitleBest ResponseYou've already chosen the best response.0
oops you are right
 2 years ago

colorful Group TitleBest ResponseYou've already chosen the best response.1
so it's nonhomogeneous I guess?
 2 years ago

colorful Group TitleBest ResponseYou've already chosen the best response.1
you could tackle it like y''+2y'=1 with undetermined coefficients or something I bet
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.