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I encountered an ODE of this form while doing research and it looks like $\frac{\mathrm{d}y}{\mathrm{d}t}+1=\frac{1}{(y+1)^2}$ 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 variable separable or by using 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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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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$\frac{dy}{dt}=\frac{1}{(y+1)^2}-1=\frac{1-(y+1)^2}{(y+1)^2}=\frac{-y(y+2)}{(y+1)^2}\implies \frac{(y+1)^2}{y(y+2)}dy=-dt$. 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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