What is the solution for e^{2t}y^'=y^2 ?

The solution for e^{2t}y^'=y^2 is y=-(2e^{2t})/(-1+2c_{1)e^{2t}}

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

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

y = - \frac{2 e ^{2 t}}{- 1 + 2 c _{1} e ^{2 t}}

Solution steps

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

Solve separable ODE:

y = - \frac{2 e ^{2 t}}{- 1 + 2 c _{1} e ^{2 t}}

y = - \frac{2 e ^{2 t}}{- 1 + 2 c _{1} e ^{2 t}}

y^{^{'}} - \frac{3 x ^{3}}{8 y ^{2}} = 0

\frac{d x}{d t} = \frac{( 8 x ln ( x ) )}{t}, x (1) = 4

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

(t^{2} - 20 t + 51) \frac{d y}{d t} = y

\frac{1}{2 x} + \frac{1}{2 y} \frac{d y}{d x} = 0

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