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Given a second order linear equation with constant coefficients, $$\frac{d^2y}{dx^2} + c\frac{dy}{dx} + dy = g(x)$$
with general solutions $$ y_g(x)=e^{-4x} (C_1\cos(3x) + C_2\sin(3x)) + 7\sin(4x)$$

Attempted solution

Notice that $y_h(x) = e^{-4x} (C_1\cos(3x) + C_2\sin(3x))$ and $y_p (x) = 7\sin(4x)$ with $y_g(x) = y_p(x) +y_h(x)$.

1) Find $c$ and $d$.

Attempt

The characteristic polynomial has roots $$ -4+3i, -4-3i, $$ hence it is $$ p(\lambda)=\lambda^2+8\lambda+25. $$ Hence the homogeoneous part is $$ y''+8y'+25y=0 $$

2) Find the forcing function, $g(x)$.

I'm a bit confused here. I believe that $e^{-4x} (C_1\cos(3x) + C_2\sin(3x))$ is the forcingg function for $x^4 e^{-4x} (C_1\cos(3x) + C_2\sin(3x))$, but we're missing the $x^4$.

3)If the system is a mass-spring system with a mass of 10kg with displacement in meters, and time in seconds, what is the spring and damping constant?

4) What is the steady state solution?

Any help appreciated. i.e. next step to be taken..

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Hint. The characteristic polynomial has roots $$ -4+3i, -4-3i, $$ hence it is $$ p(\lambda)=\lambda^2+8\lambda+25. $$ Hence the homogeoneous part is $$ y''+8y'+25y=0 $$ Next look for a paticular solution of the form $$ y_p=a\cos 4x+b\sin 4x. $$

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  • $\begingroup$ @Yigoros S. Smyrlis How do you derive the characteristic polynomial? $\endgroup$ – Anthony Peter Mar 11 '14 at 1:00
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    $\begingroup$ $e^{-4x}\cos 3x$ and $e^{-4x}\sin 3x$ correspond to the roots $\lambda_{1,2}=-4\pi 3i$. Now $\lambda_1+\lambda_2=-8$ and $\lambda_1\lambda_2=25$. $c=-$sum of the room while $d=$product of the roots. $\endgroup$ – Yiorgos S. Smyrlis Mar 11 '14 at 1:04
  • $\begingroup$ Could you give a hint regarding the next step? $\endgroup$ – Anthony Peter Mar 11 '14 at 1:13
  • $\begingroup$ S. Symrlis why are we looking for a particular solution of that form? $\endgroup$ – Anthony Peter Mar 11 '14 at 1:25

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