# How do you find a function f(x), which, when multiplied by its derivative, gives you x^3, and for which f(0) = 4?

Mar 1, 2015

The answer is: $y = \sqrt{{x}^{4} / 2 + 16}$.

This is a separable variable first-order differential equation:

$y y ' = {x}^{3} \Rightarrow y \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{3} \Rightarrow y \mathrm{dy} = {x}^{3} \mathrm{dx} \Rightarrow$

$\int y \mathrm{dy} = \int {x}^{3} \mathrm{dx} \Rightarrow {y}^{2} / 2 = {x}^{4} / 4 + c \Rightarrow y = \pm \sqrt{{x}^{4} / 2 + 2 c}$

Since the initial condition is:

$f \left(0\right) = 4$, we have to choose the positive one, and than we can find the constant $c$:

$4 = \sqrt{\frac{0}{2} + 2 c} \Rightarrow 16 = 2 c \Rightarrow c = 8$.

So the function is:

$y = \sqrt{{x}^{4} / 2 + 16}$.