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**Vamz** $\displaystyle \frac{dy}{dx} = \frac{4 x + 20 y}{5 x}$

$\displaystyle y(1)=-1$

What did I do wrong here?

$\displaystyle y'=\frac{4x}{5x}+\frac{20y}{5x}$

$\displaystyle y'-\frac{20}{5}x^{-1}y=\frac{4x}{5x}$ < Is this proper first order linear?

let $\displaystyle \eta=e^{\int\frac{20}{5x}}= e^{-4ln(x)}$

Mulitplay eta on both sides, and integrate, I get:

$\displaystyle

yx^{-4}=\frac{4}{5}\int x^{-4} dx

$

$\displaystyle

y=\frac{4x}{15}+x^4C

$

To solve for C:

$\displaystyle

-1=\frac{4}{15}+C, C= -\frac{4}{3}

$

My answer is:

$\displaystyle

y=\frac{4x}{15}-\frac{4x^4}{3}

$

which is totally wrong. What am I missing?

Thank you!