# Is partial fraction wrong way to go?

#### m3cklo

1. Homework Statement
Solve the separable equation

2. Homework Equations
dy/dx = 1-y^2

3. The Attempt at a Solution
dy/dx = 1-y^2

1/(1-y^2) dy = dx

[Partial fraction]
A/(1-y) + B/(1+y) = 1/(1-y^2)

A + Ay + B - By = 1
y^1: A - B = 0
y^0: A + B = 1
=> A=B=1/2 =>

(1/2)/(1-y) dy + (1/2)/(1+y) dy = dx

ln|1-y| + ln |1+y| = 2x + C

ln|1-y^2| = 2x + C

1-y^2 = De^(2x)

y = sqrt(1 - De^(2x))

This answer is wrong according to two different books (without explanation) that i have. The correct answer should be
y = (De^(2x) - 1)/(De^(2x) + 1)

Is partial fraction wrong way to go?
Have I made a wrong turn along the way with the algebra?
I do have big problems when i comes to solve nonlinear integrals, a tip along the way would be very appreciated!

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#### neutrino

(1/2)/(1-y) dy + (1/2)/(1+y) dy = dx

ln|1-y| + ln |1+y| = 2x + C
Check the second step.

#### m3cklo

Of course!

So obvious now, thanks a lot!!!

#### BSMSMSTMSPHD

Minus signs are your friends.

#### NTUENG

Dun like partial fractions?? Nevermind...

Hi m3cklo,

If u dun want to break the integral into partial fractions, u can use a trigo substitution instead. Let y = sin t. Then, dy = cos t dt. Simplify using the substitution. U should get sec t as the integral. Integrate it and u get ln (sec t + tan t). Convert the result back to y. U should get the same answer as the one using partial fractions.

Hope that helps.