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Popular >

y^'=e^{2y},y(0)=0

Solution

y^{^{'}} = e^{2 y}, y (0) = 0

Solution

y = - \frac{ln ( - 2 t + 1 )}{2}

Solution steps

y^{'} (t) = e^{2 y}

Solve separable ODE:

y = - \frac{ln ( - 2 t + 1 )}{2}

y = - \frac{ln ( - 2 t + 1 )}{2}

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\int 27 t^{8} e^{- t^{9}} d t

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\int (csc (x))^{2} (cot (x))^{2} d x

derivative of (3x^6+4x^3^4)

\frac{d}{d x} ((3 x^{6} + 4 x^{3})^{4})

Frequently Asked Questions (FAQ)

What is the solution for y^'=e^{2y},y(0)=0 ?
The solution for y^'=e^{2y},y(0)=0 is y=-(ln(-2t+1))/2