What is the solution for (dy)/(dx)=16\sqrt[4]{4x^4+4} ?

The solution for (dy)/(dx)=16\sqrt[4]{4x^4+4} is y=integral of 16\sqrt[4]{4x^4+4}

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(dy)/(dx)=16\sqrt[4]{4x^4+4}

\frac{d y}{d x} = 16 4 4 x^{4} + 4

y = \int 16 4 4 x^{4} + 4 d x

Solution steps

\frac{d y}{d x} = 16 4 4 x^{4} + 4

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = 16 4 4 x^{4} + 4

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int 16 4 4 x^{4} + 4 d x

y = \int 16 4 4 x^{4} + 4 d x

y^{^{''}} + 2 y^{^{'}} + 26 y = 0

2^{x}, ln (x), \frac{1}{2}, 2

x \to 4 lim (- x^{2} + 4 x)

\int 3 y e^{3 x} d x

f (x) = \frac{2}{3 x + 5}, at x = - 1

What is the solution for (dy)/(dx)=16\sqrt[4]{4x^4+4} ?
The solution for (dy)/(dx)=16\sqrt[4]{4x^4+4} is y=integral of 16\sqrt[4]{4x^4+4}