What is the solution for (t^2-1)y^'+2ty^2=0 ?

The solution for (t^2-1)y^'+2ty^2=0 is y=-1/(-ln(t^2-1)+c_{1)}

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(t^2-1)y^'+2ty^2=0

(t^{2} - 1) y^{^{'}} + 2 t y^{2} = 0

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

Solution steps

(t^{2} - 1) y^{'} (t) + 2 t y^{2} = 0

Solve separable ODE:

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

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

\int \frac{t ^{2}}{( a + b t ^{3} ) ^{2}} d t

x y^{^{'}} = 2 x^{2} y + 3 x^{3} e^{x^{2}}

\int_{4}^{9} \frac{1}{5 - x} d x

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

x \to \infty lim (x ln (1))

What is the solution for (t^2-1)y^'+2ty^2=0 ?
The solution for (t^2-1)y^'+2ty^2=0 is y=-1/(-ln(t^2-1)+c_{1)}