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- 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

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- ketz

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- zepdrix

Were you able to find the complimentary solution, \(\large y_c\) ?

- ketz

yes

- abb0t

\(y = y_h + y_c\)

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- anonymous

\[y_p = ke^{-x}cosx\]
\[dy/dp = -ke^{-x}cosx -ke^{-x}sinx\]
\[d^2y/dp^2 = 2ke^{-x}sinx\]

- anonymous

dy/dx and d^2y/dx^2 obviously lol

- anonymous

k = 1/3; my bad

- anonymous

check if the the auxillary equation is
2D^2 +4D+7=0

- 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

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

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

Because the Auxiliary Equation gave me these roots,\[\large \lambda =-1\pm\frac{\sqrt{10}}{2}i\]

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