What is the solution for yy^'-e^x=0 ?

The solution for yy^'-e^x=0 is y=sqrt(2e^x+c_{1)},y=-sqrt(2e^x+c_{1)}

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yy^'-e^x=0

y y^{^{'}} - e^{x} = 0

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

Solution steps

y y^{'} (x) - e^{x} = 0

Solve separable ODE:

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

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

y = 3 - x^{3} (0.3)

\frac{d}{d x} (lo g_{10} (2 x - 1))

\int x^{2} (x^{3} - 1)^{2} d x

\frac{d}{d x} ((5 x + 2)^{4} (2 x - 3)^{8})

y = 4 x, y = 0, x = 2

What is the solution for yy^'-e^x=0 ?
The solution for yy^'-e^x=0 is y=sqrt(2e^x+c_{1)},y=-sqrt(2e^x+c_{1)}