What is the solution for y^{''}-6y^'+9y=t^{-3}e^{3t} ?

The solution for y^{''}-6y^'+9y=t^{-3}e^{3t} is y=c_{1}e^{3t}+c_{2}te^{3t}+(e^{3t})/(2t)

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y^{''}-6y^'+9y=t^{-3}e^{3t}

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{- 3} e^{3 t}

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t}}{2 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = t^{- 3} e^{3 t}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t}}{2 t}

y = c_{1} e^{3 t} + c_{2} t e^{3 t} + \frac{e ^{3 t}}{2 t}

\frac{( 2 x y + 1 )}{y} d x + \frac{( y - x )}{y ^{2}} d y = 0

sin (sin (x))

\int (sin (x) cos^{5} (x)) d x

(4 cos (t))^{^{'}}

y = 4 x^{2}, y = 6 x^{2}, y = 6 - 5 x

What is the solution for y^{''}-6y^'+9y=t^{-3}e^{3t} ?
The solution for y^{''}-6y^'+9y=t^{-3}e^{3t} is y=c_{1}e^{3t}+c_{2}te^{3t}+(e^{3t})/(2t)