What is the solution for t^2y^{''}+3ty^'-3y= 1/t ?

The solution for t^2y^{''}+3ty^'-3y= 1/t is y=c_{1}t+(c_{2})/(t^3)-1/(4t)

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t^2y^{''}+3ty^'-3y= 1/t

t^{2} y^{^{''}} + 3 t y^{^{'}} - 3 y = \frac{1}{t}

y = c_{1} t + \frac{c _{2}}{t ^{3}} - \frac{1}{4 t}

Solution steps

t^{2} y^{''} (t) + 3 t y^{'} (t) - 3 y = \frac{1}{t}

Solve linear ODE:

y = c_{1} t + \frac{c _{2}}{t ^{3}} - \frac{1}{4 t}

y = c_{1} t + \frac{c _{2}}{t ^{3}} - \frac{1}{4 t}

\frac{2}{x - 3}

\int cos (x) (7 + 8 sin^{2} (x)) d x

6 y^{^{''}} + 40 y^{^{'}} + 54 y = 0

\frac{d}{d x} (- k x)

\int cos (x) - x sin (x) d x

What is the solution for t^2y^{''}+3ty^'-3y= 1/t ?
The solution for t^2y^{''}+3ty^'-3y= 1/t is y=c_{1}t+(c_{2})/(t^3)-1/(4t)