How to you find the general solution of $\frac{d y}{d x} = 3 x^{3}$ $dy/dx=3x^3$ ?

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Answer 1 · 2016-11-14T14:35:32

$y = \frac{3}{4} x^{4} + C$ $y = 3/4x^4+ C$

$d y = 3 x^{3} d x$ $dy = 3x^3dx$

$\int (d y) = \int (3 x^{3} d x)$ $int(dy) = int(3x^3dx)$

By the power rule, $\int (x^{n}) d x = \frac{x^{n + 1}}{n + 1}$ $int(x^n)dx = x^(n + 1)/(n + 1)$

$y = \frac{3}{4} x^{4} + C$ $y = 3/4x^4 + C$

Hopefully this helps!

How to you find the general solution of dy/dx=3x^3dydx=3x3dy/dx=3x^3?