Yes, however I believe the correct answer is y = -1/(a + x). You seem to have made a mistake in the last line.
Original post by Bilbao
dy/dx = y^2
dx/dy = 1/ y^2
∫ dx = ∫ 1/ y^2 dy
x = -1/y + A
1 =y( A - x)
is what i have done alllowed? the flipping of dy/dx to make it dx/dy ?
dx/dy = 1/ y^2
∫ dx = ∫ 1/ y^2 dy
x = -1/y + A
1 =y( A - x)
is what i have done alllowed? the flipping of dy/dx to make it dx/dy ?
Yep.
Original post by Edgemaster
Yes, however I believe the correct answer is y = -1/(a + x). You seem to have made a mistake in the last line.
Nope, he’s all good. And so are you. (The only thing that is different between the two is the sign of the arbitrary constant which is not important, but preferred to be +ve)
(edited 6 years ago)
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