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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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x \to \infty lim (\frac{2}{x} - 3)

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