What is the solution for f^'(x)=f(x)(x)^{-1}(x) ?

The solution for f^'(x)=f(x)(x)^{-1}(x) is f(x)=e^{x+c_{1}}

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f^'(x)=f(x)(x)^{-1}(x)

f^{^{'}} (x) = f (x) (x)^{- 1} (x)

f (x) = e^{x + c_{1}}

Solution steps

f^{'} (x) = f (x) x^{- 1} (x)

Solve separable ODE:

f (x) = e^{x + c_{1}}

f (x) = e^{x + c_{1}}

\frac{d y}{d x} = (\frac{y + 1}{x})

\frac{d y}{d x} = y (10000 - 50 x)

81 x^{2} + 2 y \cdot \frac{d y}{d x} = 0

y^{^{'}} + \frac{1}{2} (x - 2) y^{2} = 0, y (6) = \frac{1}{10}

y^{^{'}} - 5 y = 0, y (0) = 3

What is the solution for f^'(x)=f(x)(x)^{-1}(x) ?
The solution for f^'(x)=f(x)(x)^{-1}(x) is f(x)=e^{x+c_{1}}