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I was solving this differential equation $$ty''+2y'+ty=\cos{t}$$ with initial condition $y(0)=1$. I took Laplace on both sides and after simplifying got this differential equation in terms of $Y(s)$ $$(s^2-2)Y'(s)+4sY(s)=\frac{s}{s^2+1}+3$$ Solving this I got $$Y(s)=\frac{s^2- 3\log{(s^2+1)}+2s^3-12s}{2(s^2-2)^2}$$ What is the Laplace inverse of $Y(s)$?

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This is a linear non-homogeneous differential equation. It can be solved as

$$ y = y_h + y_p $$

such that

$$ ty''_h+2y'_h+ty_h = 0\\ ty''_p+2y'_p+ty_p = \cos t $$

To solve $y_h$ we make $y_h = \frac{e^{a t}}{t}$ and substituting

$$ (1+a^2)e^{a t} = 0 \Rightarrow y_h = \frac{C_1\sin(t)+C_2\cos(t)}{t} $$

Now substituting $y_p = C_1\sin(t)+C_2\cos(t)$ into the particular equation we get

$y_p = \frac{1}{2}\sin(t)$ and finally

$$ y = \frac{C_1\sin(t)+C_2\cos(t)}{t} + \frac{1}{2}\sin(t) $$

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  • $\begingroup$ You're missing the $1/t$ in your $y_h$ $\endgroup$ – Dylan Apr 27 '18 at 14:02
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Your Laplace equation isn't correct. Since you didn't show your work, I can't say exactly what was wrong.

Remember $\mathcal L \{ty(t)\} = -\frac{d}{ds}Y(s)$, so we have

$$ \mathcal L \{ty''\} = -\frac{d}{ds} (s^2Y - s - y'(0)) = -(2sY + s^2Y' - 1) $$

Therefore $$ -(2sY + s^2Y' - 1) + 2(sY - 1) - Y' = \frac{s}{s^2+1} $$ $$ -(s^2+1)Y' = \frac{s}{s^2+1} + 1 $$

$$ -Y' = \frac{s}{(s^2+1)^2} + \frac{1}{s^2+1} $$

You can take the inverse transform of the above to obtain

$$ t y(t) = \frac12 t\sin t + \sin t $$

$$ y(t) = \frac12 \sin t + \frac{\sin t}{t} $$

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