Is this correct?
$\displaystyle \frac{dy}{dx} = 3x^2y$
$\displaystyle \frac{1}{y} dy = 3x^2dx$
$\displaystyle \int\frac{1}{y}dy= \int(3x^2dx)$
$\displaystyle \ln|x| = x^3dx + C$
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Is this correct?
$\displaystyle \frac{dy}{dx} = 3x^2y$
$\displaystyle \frac{1}{y} dy = 3x^2dx$
$\displaystyle \int\frac{1}{y}dy= \int(3x^2dx)$
$\displaystyle \ln|x| = x^3dx + C$
Almost. Just finish up.
$\displaystyle y=e^{x^{3}+C}$
$\displaystyle y=e^{x^{3}}\cdot e^{C}$
But, $\displaystyle e^{C}$ is a constant we can call, say, K.
$\displaystyle y=Ke^{x^{3}}$
