What is the solution for (e^{(-(2x-1))}(dy))/((dx))=e^{(-y)} ?

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

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(e^{(-(2x-1))}(dy))/((dx))=e^{(-y)}

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

y = ln (\frac{1}{2} e^{2 x - 1} + c_{1})

Solution steps

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

Solve separable ODE:

y = ln (\frac{1}{2} e^{2 x - 1} + c_{1})

y = ln (\frac{1}{2} e^{2 x - 1} + c_{1})

y^{^{'}} + n^{2} y = b

\frac{d V}{d t} = 50 - \frac{7}{100} V, V = 0, t = 0

d y = e^{x + y} d x

\frac{d y}{d x} = 3 \cdot (- 4) y

y d x + (x^{2} - x) d y = 0

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