What is the solution for yy^'-16e^x=0 ?

The solution for yy^'-16e^x=0 is y=sqrt(32e^x+c_{1)},y=-sqrt(32e^x+c_{1)}

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yy^'-16e^x=0

y y^{^{'}} - 16 e^{x} = 0

y = 32 e^{x} + c_{1}, y = - 32 e^{x} + c_{1}

Solution steps

y y^{'} (x) - 16 e^{x} = 0

Solve separable ODE:

y = 32 e^{x} + c_{1}, y = - 32 e^{x} + c_{1}

y = 32 e^{x} + c_{1}, y = - 32 e^{x} + c_{1}

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What is the solution for yy^'-16e^x=0 ?
The solution for yy^'-16e^x=0 is y=sqrt(32e^x+c_{1)},y=-sqrt(32e^x+c_{1)}