Solving Separable Equations: Hi, I'm Stuck!

[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]

Welcome to PF!

Hi Aero6! Welcome to PF!

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

(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!

∫dy/y² = … ?

 

[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.