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Mathematics 19 Online
OpenStudy (anonymous):

Find the general solution of the equation dy/dt= ty^2 t/t^2+1

OpenStudy (freckles):

\[\frac{dy}{dt}=\frac{ty^2}{t^2+1} ?\]

OpenStudy (freckles):

I could be seeing this wrong because I do see another t on top also

OpenStudy (freckles):

you could do this problem by separation by variables

OpenStudy (anonymous):

Yea there another t on top. I dont know how to separate it

OpenStudy (anonymous):

Dy/dt=(ty^2 t)/(t^2+1)

OpenStudy (freckles):

\[\frac{dy}{dt}=\frac{t^2y^2}{t^2+1} \\ dy=\frac{t^2y^2}{t^2+1}dt \\ \frac{1}{y^2}dy=\frac{t^2}{t^2+1} dt \]

OpenStudy (freckles):

integrate both sides

hartnn (hartnn):

it could be \(y^2(t)\) which means square of y(t) but the approach remain same :)

OpenStudy (freckles):

hmm... That could be true. I thought the extra t was a bit odd.

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