What is the solution for (dy)/(dt)= t/(t^2y-y) ?

The solution for (dy)/(dt)= t/(t^2y-y) is y=sqrt(ln(t^2-1)+c_{1)},y=-sqrt(ln(t^2-1)+c_{1)}

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(dy)/(dt)= t/(t^2y-y)

\frac{d y}{d t} = \frac{t}{t ^{2} y - y}

y = ln (t^{2} - 1) + c_{1}, y = - ln (t^{2} - 1) + c_{1}

Solution steps

\frac{d y}{d t} = \frac{t}{t ^{2} y - y}

Solve separable ODE:

y = ln (t^{2} - 1) + c_{1}, y = - ln (t^{2} - 1) + c_{1}

y = ln (t^{2} - 1) + c_{1}, y = - ln (t^{2} - 1) + c_{1}

y^{^{'}} = (cos (x)) sin (y)

\frac{d}{d θ} (y (θ)) y (θ) = cos (e^{- 5 θ^{5}})

4 t \frac{d y}{d t} + 22 y = 0

3 y y^{^{'}} = 4, x^{3} + 7 x y = 1

y^{2} e^{2 x} d x - 4 y d y - 4 y e^{2 x} d y = 0

What is the solution for (dy)/(dt)= t/(t^2y-y) ?
The solution for (dy)/(dt)= t/(t^2y-y) is y=sqrt(ln(t^2-1)+c_{1)},y=-sqrt(ln(t^2-1)+c_{1)}