What is the solution for e^xdx=2ydy ?

The solution for e^xdx=2ydy is y=sqrt(e^x+c_{1)},y=-sqrt(e^x+c_{1)}

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e^xdx=2ydy

e^{x} d x = 2 y d y

y = e^{x} + c_{1}, y = - e^{x} + c_{1}

Solution steps

e^{x} d x = 2 y d y

Solve separable ODE:

y = e^{x} + c_{1}, y = - e^{x} + c_{1}

y = e^{x} + c_{1}, y = - e^{x} + c_{1}

\frac{d}{d y} (x + 3 y^{\frac{1}{3}})

\frac{\partial}{\partial y} (2 x y^{4})

f (x) = (x)^{(3)} e^{(3 x)}

x \to 2 + lim (e^{\frac{1 + x}{2 - x}})

y^{^{'}} + 6 y = 2 t + 3, y (0) = 1

What is the solution for e^xdx=2ydy ?
The solution for e^xdx=2ydy is y=sqrt(e^x+c_{1)},y=-sqrt(e^x+c_{1)}