How would you do this:

solve:

y^2dy - 3x^5dx = 0

the answer in the book is 2y^3 = 3x^6 + C

I am clueless how to do these types of problems. The book doesnt explain it clearly and my notes arent helping me. Thanks.

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- Aug 2nd 2007, 07:26 PMdavecs77differential equations
How would you do this:

solve:

y^2dy - 3x^5dx = 0

the answer in the book is 2y^3 = 3x^6 + C

I am clueless how to do these types of problems. The book doesnt explain it clearly and my notes arent helping me. Thanks. - Aug 2nd 2007, 07:51 PMtukeywilliams
So $\displaystyle y^2 \ dy = 3x^{5} \ dx $.

Then $\displaystyle \frac{y^3}{3} = \frac{x^6}{2} + c $

Or $\displaystyle 2y^3 = 3x^6 + 6c $

Let $\displaystyle 6c = C $ and we get

$\displaystyle 2y^3 = 3x^6 + C $

This is also separation of variables. - Aug 2nd 2007, 07:54 PMCaptainBlack
Exactly like the other one:

$\displaystyle

y^2dy - 3x^5dx = 0

$

so:

$\displaystyle

\int y^2dy - \int 3x^5dx = K

$

$\displaystyle

\frac{y^3}{3} - \frac{x^6}{2} = K

$

Now multiply through by 6:

$\displaystyle

2y^3 - 3x^6 = C

$

(the constant $\displaystyle K$ has now been replaced by the constant $\displaystyle C=6K$)

Reaarange and you have the result.

RonL