How do you solve the differential dy/dx=(x+1)/(x^2+2x-3)^2?

Jan 17, 2017

$y = - \frac{1}{2 \left({x}^{2} + 2 x - 3\right)} + c$

Explanation:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x + 1}{{x}^{2} + 2 x - 3} ^ 2$

Is a First Order Separable Differential Equation, so we can just separate the variables;

$\int \setminus \mathrm{dy} = \int \setminus \frac{x + 1}{{x}^{2} + 2 x - 3} ^ 2 \mathrm{dx}$

The LHS is immediately integrable, and for the RHS we use the substitution;

$u = {x}^{2} + 2 x - 3 \implies \frac{\mathrm{du}}{\mathrm{dx}} = 2 x + 2 = 2 \left(x + 1\right)$

Which gives:

$\setminus \setminus \setminus \setminus \setminus \int \setminus \mathrm{dy} = \int \setminus \frac{\frac{1}{2}}{u} ^ 2 \setminus \mathrm{du}$
$\therefore \int \setminus \mathrm{dy} = \frac{1}{2} \int \setminus \frac{1}{u} ^ 2 \setminus \mathrm{du}$

And now we can integrate to get;

$\setminus \setminus \setminus \setminus \setminus y = \frac{1}{2} \frac{1}{u} \left(- 1\right) + c$
$\therefore y = - \frac{1}{2 u} + c$
$\therefore y = - \frac{1}{2 \left({x}^{2} + 2 x - 3\right)} + c$