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

\frac{d}{d x} (10 x (x^{2} + 5)^{4})

x \to 0 + lim ((\frac{e ^{x} - 1}{x})^{\frac{1}{x}})

\int_{2}^{5} (x - 2)^{2} d x

\int_{0}^{5} \frac{40}{x ^{2} + 25} d x

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

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