What is the solution for t^2y^{''}+3ty^'+y=0 ?

The solution for t^2y^{''}+3ty^'+y=0 is y=(c_{1})/t+(c_{2}ln(t))/t

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t^2y^{''}+3ty^'+y=0

t^{2} y^{^{''}} + 3 t y^{^{'}} + y = 0

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

Solution steps

t^{2} y^{''} (t) + 3 t y^{'} (t) + y = 0

Solve linear ODE:

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

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

n = 1 \sum \infty 3 (\frac{2}{3})^{n - 1}

x \to 1 - lim (1 - x^{2})

\frac{d}{d x} (\frac{x ^{2} - 6 x + 7}{x ^{2} + 9})

\frac{d}{d x} (\frac{x ^{3}}{1 - x})

What is the solution for t^2y^{''}+3ty^'+y=0 ?
The solution for t^2y^{''}+3ty^'+y=0 is y=(c_{1})/t+(c_{2}ln(t))/t