What is the solution for -1/4 dv=v^2dx ?

The solution for -1/4 dv=v^2dx is x= 1/(4v)+c_{1}

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-1/4 dv=v^2dx

- \frac{1}{4} d v = v^{2} d x

x = \frac{1}{4 v} + c_{1}

Solution steps

- \frac{1}{4} d v = v^{2} d x

Let

x

be the dependent variable. Divide by

d v

- \frac{1}{4} = v^{2} \frac{d x}{d v}

Substitute

\frac{d x}{d v}

with

x^{'} (v)

- \frac{1}{4} = v^{2} x^{^{'}} (v)

Isolate

x^{'} (v) : x^{'} (v) = - \frac{1}{4 v ^{2}}

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

then

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

x = \int - \frac{1}{4 v ^{2}} d v

\int - \frac{1}{4 v ^{2}} d v = \frac{1}{4 v} + c_{1}

x = \frac{1}{4 v} + c_{1}

\frac{20}{s ( s ^{2} + 6 s + 144 )}

y = \frac{3}{x}, y = \frac{3}{x ^{2}}, x = 4

\frac{d}{d x} (e^{t a n (2 x)})

\frac{d}{d x} (\frac{2 x - 5}{3 x - 1})

x \to 2 lim (5 x^{4} - 5 x^{3} + 5)

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