What is the solution for (dy)/(dt)=((4ty^2))/(t^2+1),y(0)= 1/4 ?

The solution for (dy)/(dt)=((4ty^2))/(t^2+1),y(0)= 1/4 is y=-1/(2(ln(t^2+1)-2))

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(dy)/(dt)=((4ty^2))/(t^2+1),y(0)= 1/4

\frac{d y}{d t} = \frac{( 4 t y ^{2} )}{t ^{2} + 1}, y (0) = \frac{1}{4}

y = - \frac{1}{2 ( ln ( t ^{2} + 1 ) - 2 )}

Solution steps

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

Solve separable ODE:

y = - \frac{1}{2 ( ln ( t ^{2} + 1 ) - 2 )}

y = - \frac{1}{2 ( ln ( t ^{2} + 1 ) - 2 )}

\int x^{2} - 4 x - 5 d x

y = 9 - x^{2}, - 1 \leq x \leq 1

\int 2 x^{2} sin (\frac{x}{3}) d x

\frac{d}{d x} (9 x - 18)

\int sin (4 x) ln (sin (4 x)) d x

What is the solution for (dy)/(dt)=((4ty^2))/(t^2+1),y(0)= 1/4 ?
The solution for (dy)/(dt)=((4ty^2))/(t^2+1),y(0)= 1/4 is y=-1/(2(ln(t^2+1)-2))