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ketz
Group Title
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
 one year ago
 one year ago
ketz Group Title
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
 one year ago
 one year ago

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zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Were you able to find the complimentary solution, \(\large y_c\) ?
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
\(y = y_h + y_c\)
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.0
\[y_p = ke^{x}cosx\] \[dy/dp = ke^{x}cosx ke^{x}sinx\] \[d^2y/dp^2 = 2ke^{x}sinx\]
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.0
dy/dx and d^2y/dx^2 obviously lol
 one year ago

Euler271 Group TitleBest ResponseYou've already chosen the best response.0
k = 1/3; my bad
 one year ago

matricked Group TitleBest ResponseYou've already chosen the best response.0
check if the the auxillary equation is 2D^2 +4D+7=0
 one year ago

ketz Group TitleBest 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......????
 one year ago

zepdrix Group TitleBest 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.
 one year ago

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

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