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

The solution for y^'+2ty^2=0,y(1)= 1/5 is y= 1/(t^2+4)

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y^'+2ty^2=0,y(1)= 1/5

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

y = \frac{1}{t ^{2} + 4}

Solution steps

y^{'} (t) + 2 t y^{2} = 0

Solve separable ODE:

y = \frac{1}{t ^{2} + 4}

y = \frac{1}{t ^{2} + 4}

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

\int (cos (4 x)) d x

\frac{d}{d x} ((x^{3} + x^{2} + x + 5) ln (5 x^{2} - 7))

\frac{d}{d x} (8 x^{- 2})

\int \frac{x ^{5} - 4 x ^{3}}{x} d x

What is the solution for y^'+2ty^2=0,y(1)= 1/5 ?
The solution for y^'+2ty^2=0,y(1)= 1/5 is y= 1/(t^2+4)