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

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Answer 1 · 2017-01-17T17:33:10

$y = -1/(2(x^2+2x-3)) + c$

$dy/dx = (x+1)/(x^2+2x-3)^2$

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

$int \ dy = int \ (x+1)/(x^2+2x-3)^2 dx$

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

$u=x^2+2x-3 => (du)/dx = 2x+2 = 2(x+1)$

Which gives:

$\ \ \ \ \ int \ dy =int \ (1/2)/u^2 \ du$
$:. int \ dy =1/2int \ 1/u^2 \ du$

And now we can integrate to get;

$\ \ \ \ \ y =1/2 1/u(-1) + c$
$:. y = -1/(2u) + c$
$:. y = -1/(2(x^2+2x-3)) + c$

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