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Consider a forced, undamped spring-mass system modeled by $$x''(t) + 16x(t) = 10cos(\omega t), \omega\neq\pm4$$

Solve the above.

I am having difficulty starting this problem...

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Hint. First, solve the homogeneous equation : $$x'' + 16x = 0 $$ Then, once you found a solution $g$, apply the so-called "variant constant" method : the solutions of the equation will have the form $\lambda(t)g(t)$. This should allow you to find a suitable expression for $\lambda$.

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  • $\begingroup$ I think you mean "variation of parameters". I think it'd be easier to use the "undertermined coefficients" method since the function on the RHS is a cosine $\endgroup$ – Dylan Mar 8 '15 at 1:28
  • $\begingroup$ Yeah, "variation of parameters", I didn't know how to put it in English ^^ $\endgroup$ – Tlön Uqbar Orbis Tertius Mar 8 '15 at 7:09
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Let $x=\text{Re}(z)$, then we have the equation $$z''(t)+16z(t)=10e^{i\omega t}$$ (You could also solve this by guessing $y=Acos(\omega t)+Bsin(\omega t)$. However, the complex method is more useful for visualization purposes). Now, guess $z=Ae^{i\omega t}$ so $$z'=i\omega zAe^{i\omega t}$$ $$z''=-\omega ^2Ae^{i\omega t}$$ Plugging in we get $$-\omega ^2Ae^{i\omega t}+16Ae^{i\omega t}=10e^{i\omega t}$$ Solving for A we get $$A=\frac{10}{16-\omega ^2}$$ hence $z=\frac{10}{16-\omega^2}e^{i\omega t}$. Then, the particular solution is$$x=Re(z)=\frac{10}{16-\omega^2}\cos(\omega t)$$ Note that throughout we were using the identity $$e^{i\phi}=\cos\phi+i\sin\phi$$

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