ganeshie8
  • ganeshie8
Find a particular solution to \[y'' + 4y = f(t)\] where \(f(t)\) is given by the graph-
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ganeshie8
  • ganeshie8
|dw:1436869131827:dw|
TheViper
  • TheViper
wow XD
anonymous
  • anonymous
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UsukiDoll
  • UsukiDoll
I could've sworn that I've seen that graph before but can't remember... could be a second order ode \[r^2+4 = \] what the function of the graph is x.x
Empty
  • Empty
Well the homogeneous equation is just the wave equation which is nice, for C a complex number we have: \[y_h = c e^{i 2 x}\] and I think from here there's some Fourier transform sorta stuff we can do, since our function is essentially just an odd function raised up by a constant, so we can express it in terms of sines.
UsukiDoll
  • UsukiDoll
you guys are so good at it -_-! Teach me your secrets...another day. I am going to bed.
ganeshie8
  • ganeshie8
earlier i was hoping to shift down the graph by 1/2 units and find the coefficient \(b_n\) using usual fourier series integral as @Empty was suggesting this euler-fourier method looks neat, but is there an easy way to get real part out of \(f(t)\).. idk if it is even necessary, im just asking because im seeing for the first time...
anonymous
  • anonymous
shifting the function down by 1/2 is decomposing it into the sum of an even and an odd function, it doesn't actually simplify the resultant fourier series
anonymous
  • anonymous
it just separates it into the fourier transform of a constant and then an even periodic function, so the even and odd coefficients get separated but you have to recombine them anyways
Empty
  • Empty
Yeah you can write the function f(t) using the \(sgn\) function which just returns 1 if positive, -1 if negative, and 0 if neither as: \[f(t) = \frac{1}{2} sgn( \sin( t \pi) ) + \frac{1}{2} \] Is this what you're asking for?
anonymous
  • anonymous
hmm, consider the equation $$y''+4y=\delta(t-s)$$which gives rise to $$(\omega^2+4)Y(\omega)=e^{-i\omega s}\\Y(\omega)=\frac{e^{-i\omega s}}{\omega^2+4}$$while $$\mathcal{F}\{e^{-2|t|}\}=\frac4{\omega^2+4}\qquad\mathcal{F}\{f(t-s)\}=e^{-i\omega s}f(t)$$ so \(y(t)=\frac14 \exp(-2|t-s|)\) is the Green's function, so the solution to $$y''+4y=f(t)$$ is just $$y(t)=\frac14\int_{-\infty}^\infty e^{-2|t-s|} f(t)\,ds$$
anonymous
  • anonymous
oops, I meant \(y_s(t)=\frac14\exp(-2|t-s|)\) is the Green's function (response \(y(t)\) to an impulse at \(t=s\))
anonymous
  • anonymous
oops, I meant $$\begin{align*}y(t)&=\int_{-\infty}^\infty e^{-2|t-s|} f(s)\, ds\\&=\int_{-\infty}^t e^{2(t-s)}f(s)\, ds+\int_t^\infty e^{-2(t-s)}f(s)\,ds\\&=e^{2t}\int_{-\infty}^te^{-2s} f(s)\, ds+e^{-2t}\int_t^\infty e^{2s}f(s)\, ds\end{align*}$$now to just figure out a nice way to do those integrals
anonymous
  • anonymous
oops, i meant: $$\begin{align*}y(t)&=\int_{-\infty}^\infty e^{-2|t-s|} f(s)\, ds\\&=\int_{-\infty}^t e^{-2(t-s)}f(s)\, ds+\int_t^\infty e^{2(t-s)}f(s)\,ds\\&=e^{-2t}\int_{-\infty}^te^{2s} f(s)\, ds+e^{2t}\int_t^\infty e^{-2s}f(s)\, ds\end{align*}$$ anyways now using \(f(s)=\frac12 (\operatorname{sgn}(s)+1)\) so:$$\frac12\left(\int_{-\infty}^t e^{2s}\operatorname{sgn}(s)\,ds+\int_{-\infty}^t e^{2s}\, ds\right)$$
IrishBoy123
  • IrishBoy123
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IrishBoy123
  • IrishBoy123
**ignoring the complementary solution**, the Laplace Transform gives: \(\hat y (s^2 + 2^2)= \mathcal L\{f(t)\}\) where \(f(t) = f(t + 2)\), f(t) being the periodic repetition of \(f_1(t) = u_0(t) - u_1(t)\), period = 2 using the Rule numbers as listed in Pauls Online ...[http://tutorial.math.lamar.edu/Classes/DE/Laplace_Table.aspx] from ***Rule 34: \(\large \mathcal L \{ f_1(t) = \large \frac{\int_{0}^{2} e^{-st}} {1 - e^{-2s}} = [\frac{1}{s} - \frac{e^{-s}}{s}] . \frac{1}{1 - e^{-2s}}\) so \(\hat y(s) = \frac {1 - e^{-s}}{s(s^2 + 2^2)} . \frac{1}{1 - e^{-2s}}\) the bit at the end can be ignored as we are looking for another periodic function \(y_1(t)\) of period 2. \( y_1(t) = \mathcal L^{-1} \{ \frac {1 - e^{-s}}{s(s^2 + 2^2)} \}\) \( \mathcal L^{-1} \{ \frac {1 }{s(s^2 + 2^2)} \} = \mathcal L^{-1} \{ \frac{G(s)}{s}\}\) ***Rule 32 \( g(t) = \frac{1}{2} \ sin 2t\) \(\int_{0}^{t} \frac{1}{2} sin \ 2 \nu \ d \nu \implies \frac{1}{4} (1 - cos \ 2t)\) \( \mathcal L^{-1} \{ \frac { - e^{-s}}{s(s^2 + 2^2)} \} = \mathcal L^{-1} \{ e^{-s} F(s) \} = u_1(t)f(t-1)\) ***Rule 27 where \( f(t) = \frac{1}{4} (1 - cos \ 2t)\) so \( y_1(t) = \frac{1}{4} (1 - cos \ 2t) - \frac{1}{4} u_1(t) (1 - cos \ 2(t-1)) \) i think.

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