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Mathematics 11 Online
OpenStudy (anonymous):

separation of variables dy/dx= (1+t/1+y)^2. the whole term is raised to the second power, no parentheses on the inside terms

OpenStudy (sumi29):

Where did the t come from? Seems a bit strange.

OpenStudy (anonymous):

\[\frac{dy}{dt}=\left(\frac{1+t}{1+y}\right)^2~?\] Separating variables yields \[(1+y)^2~dy=(1+t)^2~dt\] Make substitutions: \(u=1+y~\Rightarrow~du=dy\) and \(v=1+t~\Rightarrow~dv=dt\). \[\int u^2~du=\int v^2~dv\\ \frac{1}{3}u^3=\frac{1}{3}v^3+C\\ ~~~~~~~~~~~~~\vdots\]

OpenStudy (sumi29):

@SithsAndGiggles : That is what I was gonna write, and I am almost positive that OP made a mistake in the question. Nonetheless, OP, if you reading this, confirm this solution.

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