ketz
  • ketz
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
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
Were you able to find the complimentary solution, \(\large y_c\) ?
ketz
  • ketz
yes
abb0t
  • abb0t
\(y = y_h + y_c\)

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More answers

anonymous
  • anonymous
\[y_p = ke^{-x}cosx\] \[dy/dp = -ke^{-x}cosx -ke^{-x}sinx\] \[d^2y/dp^2 = 2ke^{-x}sinx\]
anonymous
  • anonymous
dy/dx and d^2y/dx^2 obviously lol
anonymous
  • anonymous
k = 1/3; my bad
anonymous
  • anonymous
check if the the auxillary equation is 2D^2 +4D+7=0
ketz
  • ketz
but how did you choose the yp since none of e-x and cosx are found in the yc. Hence, no priority is given to either e-x or cosx......????
zepdrix
  • zepdrix
Since the right-hand 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.
zepdrix
  • zepdrix
Did your \(\large y_c\) give you this? This is what I'm coming up with, \[\large y_c=c_1e^{-x}\cos\left(\frac{\sqrt{10}}{2}x\right)+c_2e^{-x}\sin\left(\frac{\sqrt{10}}{2}x\right)\]
zepdrix
  • zepdrix
Because the Auxiliary Equation gave me these roots,\[\large \lambda =-1\pm\frac{\sqrt{10}}{2}i\]

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