1. differential 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.

2. 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.

3. Originally Posted by davecs77
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.
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(the constant $\displaystyle K$ has now been replaced by the constant $\displaystyle C=6K$)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