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

The solution for y^'=x^2e^{2y},y(-1)=-1 is y=-(ln(-(2x^3)/3+e^2-2/3))/(2)

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y^'=x^2e^{2y},y(-1)=-1

y^{^{'}} = x^{2} e^{2 y}, y (- 1) = - 1

y = - \frac{ln ( - \frac{2 x ^{3}}{3} + e ^{2} - \frac{2}{3} )}{2}

Solution steps

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

Solve separable ODE:

y = - \frac{ln ( - \frac{2 x ^{3}}{3} + e ^{2} - \frac{2}{3} )}{2}

y = - \frac{ln ( - \frac{2 x ^{3}}{3} + e ^{2} - \frac{2}{3} )}{2}

\frac{d}{d t} (3 cos (t) - 2 e^{2 t})

(- 1, 0), (0, - 2)

(x^{2} + 3 x y + y^{2}) d x - x^{2} d y = 0

cos (x), - \frac{π}{2}

\frac{5}{x}, y = 3 x, y = \frac{1}{3} x

What is the solution for y^'=x^2e^{2y},y(-1)=-1 ?
The solution for y^'=x^2e^{2y},y(-1)=-1 is y=-(ln(-(2x^3)/3+e^2-2/3))/(2)