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.(adsbygoogle = window.adsbygoogle || []).push({});

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

Here's what I got for the problem:

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

[tex]\int(1\y^2) dy = \int(1-2x)dx[/tex]

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=[tex]\sqrt{}+ or - Ce^x-x^2[/tex]

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

(-1\6)^2 = + or -([tex]\sqrt{}C[/tex])^2

C = 1\36

y=+ or - [tex]\sqrt{}1\36 e^x-x^2[/tex]

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

Thanks.

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# Separable Equations

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