Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
2((d^2)y/dx^2)+4(dy/dx)+7y=(e^x)cosx
Solve this Second order differential equation in terms of yp and yc. Explain the choice of yp. Please reply as soon as possible. Exam question...It's extremely urgent
 10 months ago
 10 months ago
2((d^2)y/dx^2)+4(dy/dx)+7y=(e^x)cosx Solve this Second order differential equation in terms of yp and yc. Explain the choice of yp. Please reply as soon as possible. Exam question...It's extremely urgent
 10 months ago
 10 months ago

This Question is Open

zepdrixBest ResponseYou've already chosen the best response.0
Were you able to find the complimentary solution, \(\large y_c\) ?
 10 months ago

Euler271Best ResponseYou've already chosen the best response.0
\[y_p = ke^{x}cosx\] \[dy/dp = ke^{x}cosx ke^{x}sinx\] \[d^2y/dp^2 = 2ke^{x}sinx\]
 10 months ago

Euler271Best ResponseYou've already chosen the best response.0
dy/dx and d^2y/dx^2 obviously lol
 10 months ago

matrickedBest ResponseYou've already chosen the best response.0
check if the the auxillary equation is 2D^2 +4D+7=0
 10 months ago

ketzBest ResponseYou've already chosen the best response.0
but how did you choose the yp since none of ex and cosx are found in the yc. Hence, no priority is given to either ex or cosx......????
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Since the righthand side is \(\large e^{x}cos x\), I think our particular solution's form will be a little bit more complicated. Since the right side has a cosine, it's telling us the particular will have the form \(\large A\cos x+B\sin x\) And the e^{x} is telling us it will have \(\large Ce^{x}\) So our y_p will be the product of these, I'll combine constants to save us some trouble. \[\large y_p=e^{x}\left(Acos x+B \sin x\right)\] And then you would take the derivative of y_p and the second derivative, \(\large y'_p=? \qquad \qquad \qquad y''_p=?\) And then use that information to plug into your y's in your equation. It's been a while since I've done this, someone can correct me if I'm wrong. But I `think` that's the y_p form that we're looking for.
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Did your \(\large y_c\) give you this? This is what I'm coming up with, \[\large y_c=c_1e^{x}\cos\left(\frac{\sqrt{10}}{2}x\right)+c_2e^{x}\sin\left(\frac{\sqrt{10}}{2}x\right)\]
 10 months ago

zepdrixBest ResponseYou've already chosen the best response.0
Because the Auxiliary Equation gave me these roots,\[\large \lambda =1\pm\frac{\sqrt{10}}{2}i\]
 10 months 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.