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}

f (x) = \frac{1}{5 + 3 x}, (1, \frac{1}{8})

- \frac{2}{s + \frac{1}{3}}

\frac{\partial}{\partial x} (sin (π (2 x - 3 y)))

x \to 0 + lim ((2 x^{2} + x)^{x})

x \to 3 lim (\frac{( x + 2 )}{x + 3})

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)}