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For $a>0$ and $\omega > 0$, show that as $t \rightarrow \infty$, the solution to the inhomogeneous differential equation $$\frac{d^2y}{dt^2}+2a\frac{dy}{dt} + y = \sin(\omega t)$$ approaches a particular solution which can be expressed in the form $$R \cos(\omega t - \phi)$$ where you should determine the amplification factor $R$ and $\tan (\phi)$.

Obviously the homogenous part doesn't matter, as it'll tend to zero.

I think I made a mistake when finding my particular solution. I tried a particular solution of $\lambda \cos(\omega t) + \mu \sin(\omega t)$ and obtained $$ \lambda = \displaystyle \frac{2a}{(1-\omega^2)^2 - 4a^2 \omega} , \mu = \displaystyle \frac{1}{1-\omega^2} + \displaystyle \frac{4a^2 \omega}{(1-\omega^2)-4a^2 \omega}$$ but this leads to horrible expressions for $R$ and $\tan (\phi)$, which I don't think are right.

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    $\begingroup$ It would be nice if you could also write down the derivatives of your trial solution, as if there is an error, it probably is located there. $\endgroup$ May 20 '17 at 13:05
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If you let $y_p = R\cos(\omega t- \phi)$ then $y_p' = -R\omega\sin(\omega t - \phi) $ and $y_p'' = -R\omega^2\cos(\omega t - \phi)$ so we have $$-R\omega^2\cos(\omega t -\phi) -2aR\omega\sin(\omega t-\phi) + R\cos(\omega t - \phi) = \sin(\omega t)$$

Expanding each term out gets you $$R\omega^2(\cos \omega t \cos \phi+\sin \omega t\sin \phi) + 2aR\omega(\sin \omega t\cos \phi -\sin \phi \cos \omega t) -R(\cos\omega t\cos \phi + \sin \omega t \sin \phi) = -\sin \omega t$$

(alternatively see LutzL's more efficient method in the comments to arrive at the next step)

So comparing the $\cos$ coefficients and noting that $R$ is non-zero gives $$\omega^2\cos \phi-2a\omega\sin \phi - \cos \phi = 0\implies \omega^2-1=2a\omega\tan \phi \Rightarrow \tan \phi = \frac{\omega^2 - 1}{2a\omega}$$

Similarly, for $\sin$ one has $$R\omega^2\sin \phi +2aR\omega\cos \phi-R\sin \phi = -1 \implies R = \frac{-1}{(\omega^2 - 1) \sin \phi + 2aw\cos \phi}$$

and you can clean out the $\phi$ dependence.

Moral: They've explicitly given you the form of the particular solution, don't use $\lambda \sin (\cdot) + \mu \cos(\cdot)$ Caveat: (there will likely be one or two minor sign errors in my answer)

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  • $\begingroup$ You could also rewrite the right side as $\sin(ωt)=\sinϕ\cos(ωt-ϕ)+\cosϕ\sin(ωt-ϕ)$ so you do not need to expand on the left side. $\endgroup$ May 20 '17 at 14:26
  • $\begingroup$ @LutzL Indeed. I initially considered that, but couldn't find a neat way to expressing $\sin (\omega t)$ in your form without pen and paper available. So I went the brute force way. I've added a mention of your approach in my answer. Thanks! $\endgroup$
    – Zain Patel
    May 20 '17 at 14:29
  • $\begingroup$ @ZainPatel hi ;) $\endgroup$ May 20 '17 at 16:21
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You should get $$ \pmatrix{1−ω^2&2aω\\-2aω&1−ω^2} \pmatrix{λ\\μ} = \pmatrix{0\\1} $$ where the matrix is of the scaling-rotation kind. Multiplying with its transpose gives $$ \Bigl((1−ω^2)^2+4a^2ω^2\Bigr)\pmatrix{λ\\μ} = \pmatrix{1−ω^2&-2aω\\2aω&1−ω^2} \pmatrix{0\\1} $$ where the result looks somewhat simpler than your formulas.


You can also think of $y=Re(z)$ where $z$ solves $$ z''+2az'+z=-ie^{iωt} $$ where you would try $z=Ce^{iωt}$ with coefficient equation $$ C(-ω^2+i2aω+1)=-i\iff C=\frac{-i}{(1-ω^2)+i2aω} $$ where you get $R$ and $-ϕ$ from the polar form of $C$, or $$ \frac1R e^{iϕ}=i((1-ω^2)+i2aω). $$

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