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The question tells us that the functions $x(t)$ and $y(t)$ satisfy $dx/dt=-x^2y$ and $dy/dt= -xy^2$ when $t=0$, $x=1$, and $y=2$. I have already worked out that $dy/dx= (xy^2)/(x^2y)$ and that $y=2x$. Now given that $dx/dt=-2x^3$ deduce a formula for $x$ as a function of $t$.

Any help would be much appreciated

Morten
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Christie
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  • Do you know how to solve the differential equation $x'(t) = -2x^3, x(0) = 1$? – BaronVT Dec 04 '14 at 18:42
  • No I don't think so – Christie Dec 04 '14 at 18:44
  • Does "separable differential equation" sound familiar: you can write $\dfrac{dx}{dt} = -2x^3$, leading to $\dfrac{dx}{-2 x^3} = dt$, and then you can integrate (see charlotte's solution below) – BaronVT Dec 04 '14 at 18:48
  • Well I probably do I just can't think of how to do it – Christie Dec 04 '14 at 18:49
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    Ohhh that makes sense!! Thank you!! – Christie Dec 04 '14 at 18:49
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  • And you will get points yourself for accepting an answer, @RachelBinnie Everyone wins! – BaronVT Dec 04 '14 at 19:16

1 Answers1

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You're almost there. You just have to separate the variables -

$$\frac{dx}{dt}=-2x^3$$ $$-\frac{dx}{2x^3}=dt$$ $$-\int\frac{dx}{2x^3}= t+C$$ $$\frac{1}{4x^2}=t+C$$ $$x=\pm\frac{1}{2\sqrt{t+C}}$$

$$1 = \pm\frac{1}{2\sqrt{C}}$$

Assuming $C\in\mathbb{R}$, this means $C=\frac{1}{4}$ and only the positive solution is admitted. Then $$x = \frac{1}{\sqrt{4t+1}}$$

charlotte
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