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A nonlinear ODE (Part II)

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We could have directly solved the non-linear differential equation of the previous post exactly. This is now what we’re going to do, and we’ll be able to observe the equivalence between the exact solution and the corresponding perturbative series.

$\displaystyle y' + y + y^2 = 0$

$\displaystyle y' = -(y + y^2)$

$\displaystyle \frac{1}{y + y^2}\,dy = -\,dx$

$\displaystyle \int \left(\frac{1}{y+y^2}\right) dy = - \int dx$

$\displaystyle \int \frac{1}{y(1 + y)}\,dy = - \int dx$

$\displaystyle \int \left(\frac{1}{y} - \frac{1}{1 + y}\right) dy = - \int dx$

$\displaystyle \ln(y) - \ln(1 + y) = - x + C$

$\displaystyle \frac{y}{y+1} = k e^{-x}$

$\displaystyle \frac{y}{y + 1} - \frac{y+1}{y+1} = k e^{-x} -1$

$\displaystyle - \frac{1}{y + 1} = k e^{-x} -1$

$\displaystyle -\frac{1}{k e^{-x}-1} = y + 1$

$\displaystyle y = \frac{-k e^{-x}}{k e^{-x}-1}$

Using the initial condition $y(0) = 1$ we have $k=\frac{1}{2}$ . In a similar way, we can show that the exact solution to the perturbed differential equation :