- #1
checkmatechamp
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I'm a little rusty with partial fractions, and I can't seem to find my error once I get up to that point.
dy/dx = (y^2 - 1) / x
Cross-mutliply
x dy = (y^2 - 1) dx
Divide by the appropriate terms
dy / (y^2 - 1) = dx / x
So I'm integrating 1 / (y^2 - 1) dy and 1/x dy
Obviously, the integral of 1/x is ln|x| + c, but I'm having trouble with integrating the y terms. This is what I did so far.
A / (y + 1) + B/ (y - 1) = 1 / (y^2 - 1)
Ay - A + By + B = 1
Combine the y terms and constant terms
A + B = 0 (Since there's no y term)
-A + B = 1
2B = 1
B = 0.5
So I integrate 0.5 / (y - 1) and get 0.5*ln|y - 1|
So that means 0.5*ln|y - 1| = 1/|x| + c
ln|y - 1| = 2/x + c
y - 1 = e^(2/x + c)
y - 1 = e^(2/x) * e^c (which is just a constant)
y = ce^(2/x) + 1
But the thing is that I checked online to find out how to integrate the y term, and found this: http://calc101.com/partial_1.html
My question is: Why does 1 / (y + 1)(y - 1) get split up into 1/(2y - 2) and 1/(2y + 2)? Also, why are they subtracting those 2 terms rather than adding them?
Homework Statement
dy/dx = (y^2 - 1) / x
Homework Equations
The Attempt at a Solution
Cross-mutliply
x dy = (y^2 - 1) dx
Divide by the appropriate terms
dy / (y^2 - 1) = dx / x
So I'm integrating 1 / (y^2 - 1) dy and 1/x dy
Obviously, the integral of 1/x is ln|x| + c, but I'm having trouble with integrating the y terms. This is what I did so far.
A / (y + 1) + B/ (y - 1) = 1 / (y^2 - 1)
Ay - A + By + B = 1
Combine the y terms and constant terms
A + B = 0 (Since there's no y term)
-A + B = 1
2B = 1
B = 0.5
So I integrate 0.5 / (y - 1) and get 0.5*ln|y - 1|
So that means 0.5*ln|y - 1| = 1/|x| + c
ln|y - 1| = 2/x + c
y - 1 = e^(2/x + c)
y - 1 = e^(2/x) * e^c (which is just a constant)
y = ce^(2/x) + 1
But the thing is that I checked online to find out how to integrate the y term, and found this: http://calc101.com/partial_1.html
My question is: Why does 1 / (y + 1)(y - 1) get split up into 1/(2y - 2) and 1/(2y + 2)? Also, why are they subtracting those 2 terms rather than adding them?