Question

obtain the general solution:
(1-x)y'=y^2

the ans. is 1= ylnC(1-X)

but when i answered it, it was 1 = -ylnC(1-x)

how can i remove then negative(-) sign?

Answer 1

You got the sign right on the y-side, but you forgot your chain rule on the x-side.

BTW \(\int \dfrac{1}{1-x}\;dx = -ln|1-x| + C \ne -ln(1-x) + C\)

Answer 2

uh, whats the difference b/w them?

Answer 3

Absolute values and not absolute values? For the logarithm function, you must KNOW the argument is positive.

Answer 4

oh, so to make the natural logarithm positive, you make (1-x) as absolute values therefore |1-x| right?

Answer 5

This is a common point of confusion.

\(\dfrac{d}{dx}ln(x) = \dfrac{1}{x}\)

However, \(\int \dfrac{1}{x}\;dx = ln|x| + C\)

When we did the derivative, we already had a valid logarithm function, so we already know x > 0.

When we did the integral, we don't know anything about x so we have to force it to be positive in order to introduce the logarithm.

Making a little bit of sense?

If you KNOW x is positive, then you can just forget about it.

Answer 6

oh, thank you!