Solve $\frac{dy}{dt} = y$ with inital condition y(0) = -1, t(0) = 0
attempt $\int \frac{dy}{y} = \int dt$
$lny = t + c$
$lny = t + lnc$ $y = ce^t$ $y(0) = -1 = c$
so $y = t =1$
Solve $\frac{dy}{dt} = y$ with inital condition y(0) = -1, t(0) = 0
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ordinary-differential-equations
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0Soooooooooooooo.... wheres the question/ – 2017-02-27
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0Actually, since $c=-1$, it follows that $y=ce^t=-e^t$ – 2017-02-27
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0@BrevanEllefsen I think many users take "Solve ..." as the question, though it has no actual question marks ;) – 2017-02-27
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0@SimplyBeautifulArt fair enough. I just avoid helping until the OP actually expresses what their question is. The error in the work is clear, but I wanted the *OP to realize this* – 2017-02-27
1 Answers
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I am not sure what $t(0)=0$ means, but except that the answer would be $$\frac{dy}{dx}\ =\ y$$ $$\int\frac{dy}y\ =\ \int dt$$ $$\ln|y|\ =\ t+C$$ $$|y|\ =\ e^{t+c}$$ Since $\ y(0)=-1$, $$|y(0)|=|-1|=e^C$$ $$C=0$$ Thus, $$y=-e^t$$