What is the solution for y^'*ln(y)=y ?

The solution for y^'*ln(y)=y is y=e^{sqrt(2t+c_{1)}},y=e^{-sqrt(2t+c_{1)}}

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y^'*ln(y)=y

y^{^{'}} \cdot ln (y) = y

y = e^{2 t + c_{1}}, y = e^{- 2 t + c_{1}}

Solution steps

y^{'} (t) \cdot ln (y) = y

Solve separable ODE:

y = e^{2 t + c_{1}}, y = e^{- 2 t + c_{1}}

y = e^{2 t + c_{1}}, y = e^{- 2 t + c_{1}}

x = \frac{y}{y ^{^{'}} + y}

20 (\frac{d v}{d t}) = - (20) (32) - 0.0025 v

\frac{d y}{d x} \cdot (1 + e^{x}) = e^{x - y}

y^{^{'}} = - 9.8 - \frac{0.0025}{4} y

y^{^{'}} = y tan (x)

What is the solution for y^'*ln(y)=y ?
The solution for y^'*ln(y)=y is y=e^{sqrt(2t+c_{1)}},y=e^{-sqrt(2t+c_{1)}}