What is the solution for t^2y^{''}+7ty^'+5y=t ?

The solution for t^2y^{''}+7ty^'+5y=t is y=(c_{1})/t+(c_{2})/(t^5)+t/(12)

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t^2y^{''}+7ty^'+5y=t

t^{2} y^{^{''}} + 7 t y^{^{'}} + 5 y = t

y = \frac{c _{1}}{t} + \frac{c _{2}}{t ^{5}} + \frac{t}{12}

Solution steps

t^{2} y^{''} (t) + 7 t y^{'} (t) + 5 y = t

Solve linear ODE:

y = \frac{c _{1}}{t} + \frac{c _{2}}{t ^{5}} + \frac{t}{12}

y = \frac{c _{1}}{t} + \frac{c _{2}}{t ^{5}} + \frac{t}{12}

\frac{d}{d x} (\frac{2}{5} x^{15})

t^{2} cos (2 t + 3 0^{\circ})

f (x) = 5 x sin (π x)

y^{^{''}} - 2 y^{^{'}} + y = e^{x} tan^{- 1} (x)

x \to - 2 lim ([f (x) + g (x)])

What is the solution for t^2y^{''}+7ty^'+5y=t ?
The solution for t^2y^{''}+7ty^'+5y=t is y=(c_{1})/t+(c_{2})/(t^5)+t/(12)