Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

3 Answers 3

$$ \frac{dy}{dt} = \frac{1-(y+1)^2}{(y+1)^2}. $$ 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.

share|improve this answer
    
Ah! Thank you Siminore for the suggestion! Can't believe I didn't spot that! –  Joe Jul 11 '12 at 16:43

You can separate the variables: $$-\left(1+\frac{1}{y(y+2)}\right)dy=dt$$ Split into partial fractions, integrate both sides! $$-\left(1+\frac1{2y}-\frac1{2(y+2)}\right)dy=dt$$ $$-y+\frac12\log\frac{y+2}y=t+c$$ $$\sqrt\frac {y+2}ye^{-y}=ke^t$$

share|improve this answer
    
There should be $y+2$ in place of $y-2$. –  Aang Jul 11 '12 at 16:34
    
Hey Aneesh karthik C, there seems to be minus missing in line two! However, thanks for the suggestion! –  Joe Jul 11 '12 at 17:01
    
oops. right! Thank you! –  Host-website-on-iPage Jul 11 '12 at 17:07

$\frac{dy}{dt}=\frac{1}{(y+1)^2}-1=\frac{1-(y+1)^2}{(y+1)^2}=\frac{-y(y+2)}{(y+1)^2}\implies \frac{(y+1)^2}{y(y+2)}dy=-dt$. Integrating both sides gives $\int \frac{(y+1)^2}{y(y+2)}dy=-\int dt=\int \frac{y^2+2y+1}{y^2+2y}dy=-\int dt\implies \int (1+\frac{1}{y(y+2)})dy=-\int dt\implies y+ \frac{1}{2}\int ((1/y)-1/(y+2))dy=-t+c\implies y+\frac{1}{2}\ln{\frac{y}{y+2}}=-t+c$.

share|improve this answer
    
Hey avator, shouldn't there be a minus in the first line? Otherwise, thanks for the solution! –  Joe Jul 11 '12 at 16:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.