What is the solution for (y)dx-4(x+y^6)dy=0 ?

The solution for (y)dx-4(x+y^6)dy=0 is y=c_{1}

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(y)dx-4(x+y^6)dy=0

(y) d x - 4 (x + y^{6}) d y = 0

y = c_{1}

Solution steps

(y) d x - 4 (x + y^{6}) d y = 0

Let

y

be the dependent variable. Divide by

d x

\frac{d y}{d x} - 4 (x + y^{6}) \frac{d y}{d x} = 0

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) - 4 (x + y^{6}) y^{^{'}} (x) = 0

Isolate

y^{'} (x) : y^{'} (x) = 0

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

then

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

y = \int 0 d x

\int 0 d x = c_{1}

y = c_{1}

f (x) = 3 - 4 x^{2}, (4, - 61)

5 cos (t)

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

y^{^{''}} + 16 y = - 16 sec^{2} (4 x)

\frac{\partial}{\partial x} (\frac{c + 1}{c + x})

What is the solution for (y)dx-4(x+y^6)dy=0 ?
The solution for (y)dx-4(x+y^6)dy=0 is y=c_{1}