What is the solution for y^'=((3t^2-e^t))/((2y-5)),y(0)=1 ?

The solution for y^'=((3t^2-e^t))/((2y-5)),y(0)=1 is y=(5-sqrt(4t^3-4e^t+13))/2

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y^'=((3t^2-e^t))/((2y-5)),y(0)=1

y^{^{'}} = \frac{( 3 t ^{2} - e ^{t} )}{( 2 y - 5 )}, y (0) = 1

y = \frac{5 - 4 t ^{3} - 4 e ^{t} + 13}{2}

Solution steps

y^{'} (t) = \frac{( 3 t ^{2} - e ^{t} )}{( 2 y - 5 )}

Solve separable ODE:

y = \frac{5 - 4 t ^{3} - 4 e ^{t} + 13}{2}

y = \frac{5 - 4 t ^{3} - 4 e ^{t} + 13}{2}

n = 1 \sum \infty ln (n + 2)

x \to 2 - lim (ln (x^{2} (3 - x)))

x^{2}, 2 - x^{2}

x \to 0 lim (\frac{1}{2})

\int_{0}^{1} x (1 - x)^{8} d x

What is the solution for y^'=((3t^2-e^t))/((2y-5)),y(0)=1 ?
The solution for y^'=((3t^2-e^t))/((2y-5)),y(0)=1 is y=(5-sqrt(4t^3-4e^t+13))/2