Originally Posted by

**Paymemoney** ok i have tried it again my answer does not match the book's answers.

$\displaystyle \frac{dy}{dx} - \frac{4y}{x} = x^4$

$\displaystyle e^{\int \ln(x^4) \, dx} = x^4$ Mr F says: I have made some formatting and notational changes to this line without changing its content. My request to you is this: Please explain where it has come from, noting that, as has previously been pointed out to you, p(x) = -4/x.

$\displaystyle x^4(\frac{dy}{dx} - \frac{4y}{x}) = x^8$

$\displaystyle (\frac{d(x^4y)}{dx} = x^8$

$\displaystyle x^4y = \int x^8$

$\displaystyle x^4y = \frac{x^9}{9} + C$

$\displaystyle y = \frac{x^5}{9} + \frac{C}{x^4}$