# Math Help - differential equations

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 $y^2 \ dy = 3x^{5} \ dx$.

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

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

Let $6c = C$ and we get

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

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:

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

so:

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

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

Now multiply through by 6:

$
2y^3 - 3x^6 = C
$

(the constant $K$ has now been replaced by the constant $C=6K$)

Reaarange and you have the result.

RonL