## 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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1. zepdrix Group Title

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

2. ketz Group Title

yes

3. abb0t Group Title

$$y = y_h + y_c$$

4. Euler271 Group Title

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

5. Euler271 Group Title

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

6. Euler271 Group Title

7. matricked Group Title

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

8. ketz Group Title

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......????

9. zepdrix Group Title

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.

10. zepdrix Group Title

Did your $$\large y_c$$ give you this? This is what I'm coming up with, $\large y_c=c_1e^{-x}\cos\left(\frac{\sqrt{10}}{2}x\right)+c_2e^{-x}\sin\left(\frac{\sqrt{10}}{2}x\right)$

11. zepdrix Group Title

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