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

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

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1/t y^'= y/(t^2)-1/y

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\frac{\partial}{\partial y} (5 y - 2 x)

2 y^{^{''}} + 8 y^{^{'}} + 80 y = 100 cos (8 t)

(2 x - 1) d x + (7 y + 5) d y = 0

\frac{d}{d x} (2 e^{x - 1})

\int tan (x) \cdot sec^{2} (x) d x

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