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Quiz-1 / Re: Q1: TUT0301
« on: September 29, 2018, 03:13:34 PM »
$$
y' - 2y = e^{2t}\\\mu (x) = e^{\int-2\ dt} = e^{-2t}\\\frac{d}{dt} (e^{-2t}y) = e^{-2t}y' -2e^{-2t} y = e^{-2t}e^{2t}= 1\\ \text{integral on both side gives:}\\ e^{-2t}y = t + C\\y = te^{2t} + C{e^{2t}}\\y(0) = 0 + C = 2 \implies C = 2\\\text{thus} \ \ y =e^{2t}(t+2)\\\text{and} \ \ t \to \infty \implies y \to \infty
$$
y' - 2y = e^{2t}\\\mu (x) = e^{\int-2\ dt} = e^{-2t}\\\frac{d}{dt} (e^{-2t}y) = e^{-2t}y' -2e^{-2t} y = e^{-2t}e^{2t}= 1\\ \text{integral on both side gives:}\\ e^{-2t}y = t + C\\y = te^{2t} + C{e^{2t}}\\y(0) = 0 + C = 2 \implies C = 2\\\text{thus} \ \ y =e^{2t}(t+2)\\\text{and} \ \ t \to \infty \implies y \to \infty
$$