Separable Equations

[Aero6]

Hi I've been working on this problem repeatedly and thought I understood how to solve separable equations problems but I keep getting the wrong answer.

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

Here's what I got for the problem:

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

\int(1\y^2) dy = \int(1-2x)dx

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

e^ln|y^2| =e^ x-x^2 +C , e^C = C

y^2= + e^x-x^2 +C or y^2 = -e^x-x^2 +C

y=\sqrt{}+ or - Ce^x-x^2

-1/6=+ or - \sqrt{}Ce^ 0-0^2

(-1\6)^2 = + or -(\sqrt{}C)^2

C = 1\36

y=+ or - \sqrt{}1\36 e^x-x^2


Also, I'm not sure what to do when the problem tells you to determine the interval in which the solution is valid.

Thanks.

[tiny-tim]

Hi Aero6! Welcome to PF! :smile:

(have a squared: ² and an integral: ∫ :smile:)

(and fractions in LaTeX are \frac{}{})

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

Here's what I got for the problem:

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

\int\frac{1}{y^2} dy = \int(1-2x)dx

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

erm … you got too excited at seeing the 1/ in the ∫ … they're not all logs! :wink:

∫dy/y² = … ? :smile:

[HallsofIvy]

Hi I've been working on this problem repeatedly and thought I understood how to solve separable equations problems but I keep getting the wrong answer.

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

Here's what I got for the problem:

y'y^2 = (1-2x)
Shouldn't it be y'/y^2= 1-2x ?

\int(1/y^2) dy = \int(1-2x)dx
Yes, it was!

ln | y^2| = x-x^2 + C
No, the integral of y-2 is -y-1

e^ln|y^2| =e^ x-x^2 +C , e^C = C

y^2= + e^x-x^2 +C or y^2 = -e^x-x^2 +C

y=\sqrt{}+ or - Ce^x-x^2

-1/6=+ or - \sqrt{}Ce^ 0-0^2

(-1\6)^2 = + or -(\sqrt{}C)^2

C = 1\36

y=+ or - \sqrt{}1\36 e^x-x^2


Also, I'm not sure what to do when the problem tells you to determine the interval in which the solution is valid.

Thanks.