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I encountered an ODE of this form while doing research and it looks like $\frac{dy}{dt}+1=\frac1{(y+1)^2}$ How would one go back solving such an equation? Wolfram Alpha gave me a solution but I would still like to know the idea behind getting one!

It doesn't look to me as though we can solve it through separation or by integrating factors.

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$ \frac{dy}{dt} = \frac{1-(y+1)^2}{(y+1)^2}. $ The right-hand side is a function of $y$ only, so you can separate variables: $ \frac{(y+1)^2}{1-(y+1)^2}dy = dt, $ and both sides can be integrated elementarily.

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    Ah! Thank you Siminore for the suggestion! Can't believe I didn't spot that!2012-07-11
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$\frac{dy}{dt}=\frac{1}{(y+1)^2}-1=\frac{1-(y+1)^2}{(y+1)^2}=\frac{-y(y+2)}{(y+1)^2}\implies \frac{(y+1)^2}{y(y+2)}dy=-dt$. Integrating both sides gives $\int \frac{(y+1)^2}{y(y+2)}dy=-\int dt=\int \frac{y^2+2y+1}{y^2+2y}dy=-\int dt\implies \int (1+\frac{1}{y(y+2)})dy=-\int dt\implies y+ \frac{1}{2}\int ((1/y)-1/(y+2))dy=-t+c\implies y+\frac{1}{2}\ln{\frac{y}{y+2}}=-t+c$.

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    Hey avator, shouldn't there be a minus in the first line? Otherwise, thanks for the solution!2012-07-11
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You can separate the variables: $-\left(1+\frac{1}{y(y+2)}\right)dy=dt$ Split into partial fractions, integrate both sides! $-\left(1+\frac1{2y}-\frac1{2(y+2)}\right)dy=dt$ $-y+\frac12\log\frac{y+2}y=t+c$ $\sqrt\frac {y+2}ye^{-y}=ke^t$

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    oops. right! Thank you!2012-07-11