What is the solution for y^'=(x+1)/(e^{2y)} ?

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

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y^'=(x+1)/(e^{2y)}

y^{^{'}} = \frac{x + 1}{e ^{2 y}}

y = \frac{ln ( x ^{2} + 2 x + c _{1} )}{2}

Solution steps

y^{'} (x) = \frac{x + 1}{e ^{2 y}}

Solve separable ODE:

y = \frac{ln ( x ^{2} + 2 x + c _{1} )}{2}

y = \frac{ln ( x ^{2} + 2 x + c _{1} )}{2}

5 x cos (5 y) d y - sin (5 y) d x = 0

x y d x + 1 + x^{2} d y = 0

y^{^{'}} + y cos (x) = 3 cos (x)

\frac{d y}{d x} + \frac{e ^{y^{2}}}{x ^{2} y} = 0, y (1) = 0

x \frac{d y}{d x} = y x - x y cos (x)

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