What is the solution for 1/5 (dy)/(dx)=(1/5)^2-1/5 ?

The solution for 1/5 (dy)/(dx)=(1/5)^2-1/5 is y=-4/5 x+c_{1}

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1/5 (dy)/(dx)=(1/5)^2-1/5

\frac{1}{5} \frac{d y}{d x} = (\frac{1}{5})^{2} - \frac{1}{5}

y = - \frac{4}{5} x + c_{1}

Solution steps

\frac{1}{5} \frac{d y}{d x} = (\frac{1}{5})^{2} - \frac{1}{5}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

\frac{1}{5} y^{^{'}} (x) = (\frac{1}{5})^{2} - \frac{1}{5}

Isolate

y^{'} (x) : y^{'} (x) = - \frac{4}{5}

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

then

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

y = \int - \frac{4}{5} d x

\int - \frac{4}{5} d x = - \frac{4}{5} x + c_{1}

y = - \frac{4}{5} x + c_{1}

\frac{d}{d x} (y + x cos (y))

\frac{5 x + 5}{x ^{2} - 4}

\int \frac{e ^{l n (x)}}{x ^{2}} d x

\frac{d}{d x} (\frac{x ^{2} - 5 x}{x + 1})

x \to 4 lim (x 20 - x^{2})

What is the solution for 1/5 (dy)/(dx)=(1/5)^2-1/5 ?
The solution for 1/5 (dy)/(dx)=(1/5)^2-1/5 is y=-4/5 x+c_{1}