What is the solution for yy^'-8e^x=0 ?

The solution for yy^'-8e^x=0 is y=sqrt(16e^x+c_{1)},y=-sqrt(16e^x+c_{1)}

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yy^'-8e^x=0

y y^{^{'}} - 8 e^{x} = 0

y = 16 e^{x} + c_{1}, y = - 16 e^{x} + c_{1}

Solution steps

y y^{'} (x) - 8 e^{x} = 0

Solve separable ODE:

y = 16 e^{x} + c_{1}, y = - 16 e^{x} + c_{1}

y = 16 e^{x} + c_{1}, y = - 16 e^{x} + c_{1}

ln (2 x + 5)

sin (x^{2})

\int e^{2 t} t d t

f (x) = 3 x^{2} - 5 x - 1 + e^{- x}, at x = 0

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What is the solution for yy^'-8e^x=0 ?
The solution for yy^'-8e^x=0 is y=sqrt(16e^x+c_{1)},y=-sqrt(16e^x+c_{1)}