What is the solution for yy^'=e^{2x} ?

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

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yy^'=e^{2x}

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

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

Solution steps

y y^{'} (x) = e^{2 x}

Solve separable ODE:

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

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

\frac{d y}{d x} = 56 y x^{7}

\frac{d y}{d x} = \frac{x ( y + 2 ) ^{2}}{e ^{- x^{2}}}

\frac{d y}{d x} 4 y^{2} = \frac{3 x - 15}{3 x + 15}

\frac{d y}{d x} x = 2 y

y^{^{'}} = 2.25 y

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