find particular solution y=f(x

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find particular solution y=f(x

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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its easy
\[\frac{ dy }{ dx }=6x^2-x^2y\]
\[\Large \frac{dy}{dx}=(6-y)x^2 \] Or \[ \Large \frac{1}{6-y}dy=x^2dx \] For \(y(x) \neq 6\)

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Other answers:

\[\int\limits_{}^{}\frac{ 1 }{ 6-y }dy=x^2dx\]
Integrate both sides.
\[-lny=2x^3+c\]
\[\Large - \ln |6-y(x)|=\frac{1}{3}x^3+C\prime  \]
they have given f(-1)=2
no calc
calculator
You know you have to solve it for y(x) if you can, of course - you could apply the initial conditions just now and solve for the constant. But I strongly recommend you to get the equation in explicit form if possible.
i got \[y(x)=ce ^{(x^3/x)}+6\]
Almost, try again. You should end up with: \[\Large y(x)=6-Ce^{-\frac{1}{3}x^3} \] As you can see, I said above that \(y(x) \neq 6\) which was important, otherwise I would have divided the differential equation through 0. This equation supports that statement, this equation only becomes 6 in the limit as x approaches infinity.
|dw:1360441531205:dw|
|dw:1360441583641:dw|
i alway get these
Consider this equation: \[\Large e^{\ln2} = 2 \] but: \[\Large e^{-\ln2} = -2 ????? \] Rather view it like that: \[\Large e^{-\ln2}=\left(e^{\ln2}\right)^{-1}=\frac{1}{2} \]
okay
and then you should get the same expression as I do.

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