What is the solution for (dy)/(dx)=(2^{4x-y}) ?

The solution for (dy)/(dx)=(2^{4x-y}) is y=(ln(2^{4x-2}+c_{1}))/(ln(2))

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(dy)/(dx)=(2^{4x-y})

\frac{d y}{d x} = (2^{4 x - y})

y = \frac{ln ( 2 ^{4 x - 2} + c _{1} )}{ln ( 2 )}

Solution steps

\frac{d y}{d x} = (2^{4 x - y})

Solve separable ODE:

y = \frac{ln ( 2 ^{4 x - 2} + c _{1} )}{ln ( 2 )}

y = \frac{ln ( 2 ^{4 x - 2} + c _{1} )}{ln ( 2 )}

\frac{d}{d x} ((x^{3} - x) e^{x})

\frac{d}{d x} (x^{2} + x^{- 2})

\frac{\partial}{\partial x} (e^{2 y} cos (- 2 x))

\frac{d}{d x} (12 cos (x))

x \to \infty lim (x^{x})

What is the solution for (dy)/(dx)=(2^{4x-y}) ?
The solution for (dy)/(dx)=(2^{4x-y}) is y=(ln(2^{4x-2}+c_{1}))/(ln(2))