What is the solution for y^'=-1/5 y^2,y(0)=0.25 ?

The solution for y^'=-1/5 y^2,y(0)=0.25 is y= 5/(t+20)

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y^'=-1/5 y^2,y(0)=0.25

y^{^{'}} = - \frac{1}{5} y^{2}, y (0) = 0.25

y = \frac{5}{t + 20}

Solution steps

y^{'} (t) = - \frac{1}{5} y^{2}

Solve separable ODE:

y = \frac{5}{t + 20}

y = \frac{5}{t + 20}

4 \frac{d y}{d x} = \frac{e ^{y} sin ^{2} ( x )}{y sec ( x )}

\frac{d y}{d x} + x \cdot y = x \cdot y^{2}

y^{^{'}} = 0.1044 \cdot y, y (0) = 100000

\frac{d y}{d x} = \frac{8 y}{9 x}

What is the solution for y^'=-1/5 y^2,y(0)=0.25 ?
The solution for y^'=-1/5 y^2,y(0)=0.25 is y= 5/(t+20)