Question

Using the differential equation dy/dt = t^2/(2y + 2t^3y), find the domain of the solution with initial condition y(0) = -2.
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Answer 1

Please don't mass tag people.

Answer 2

I'm sorry I can not help you.

Answer 3

\[\large \frac{dy}{dt}=\frac{1}{y} \cdot \frac{t^2}{2+2t^3} \implies y dy = \frac{t^2}{2+2t^3}dt \]
Solve both sides by integration and then study the domain separately. Left hand side is primitive, right hand side is a basic \(u-\) Substitution.

Answer 4

Ok so I have y(t) = -sqrt(ln(abs(1+t^3))+4), now what's the domain of this solution?

Answer 5

I don't get the exact same answer, it's missing a 1/3 in front of the logarithmic expression, however, how do you study a domain? You study what values of \(x\) you can plugin for which the function is well defined. In your case, you do the same but for \(t\)

Answer 6

hint, you can't take the square root of a negative number in \(\mathbb{R}\)

Answer 7

My main question is: Is the lower boundary (-1+e^(-4))^(1/3) included in the domain, and why?

Answer 8

I would have said the lower boundary is \[\large \left(e^{-12}-1 \right)^{\frac{1}{3}} \]

Answer 9

But my solution is different from yours, with the solution you have mentioned above it would work out, because you have:
\[\large y(t)=- \sqrt{ \log(t^3+1)+4} \] So your square root mustn't be zero: that would imply that:
\[\large \sqrt{\log (t^3+1)+4}=0 \implies \log (t^3+1)=-4 \implies t^3= e^{-4}-1\]from which you can take the 3rd square root to get your desired lower bound, however, let me mention again that I got a different result.

Answer 10

$$\frac{dy}{dt}=\frac1{2y}\cdot\frac{t^2}{1 + t^3}\\2y\frac{dy}{dt}=\frac{t^2}{1+t^3}\\2y\frac{dy}{dt}=\frac13\frac{3t^2}{1+t^3}\\\int 2y\frac{dy}{dt}dt=\int\frac13\frac{3t^2}{1+t^3}dt\\y^2=\frac13\log(1+t^3)+C$$

Answer 11

since \(y(0)<0\) we pick:$$y(t)=-\sqrt{\frac13\log(1+t^3)+C}$$

Answer 12

so$$y(0)=-\sqrt{\frac13\log(1)+C}\\-2=-\sqrt{C}\\C=4$$

Answer 13

My biggest question was what is the domain of the initial condition? Honestly, I already knew what you gave me :P :P :P @oldrin.bataku

Answer 14

okay... then think logically. we've found our solution$$y(t)=-\sqrt{\frac13\log(1+t^3)+4}$$and we know the domain of \(\sqrt{x}\) is \(x\ge0\) i.e.$$\frac13\log(1+t^3)+4\ge0\\\frac13\log(1+t^3)\ge-4\\\log(1+t^3)\ge-12\\1+t^3\ge e^{-12}\\t^3\ge e^{-12}-1\\t\ge\sqrt[3]{e^{-12}-1}$$

Answer 15

note the domain of \(\log x\) is \(x\ge0\) therefore $$1+t^3\ge0\\t^3\ge-1\\t\ge-1$$but we recognize this bound is redundant given the above