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

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

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y^{''}-6y^'+9y=((e^{3t}))/(t^2)

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{( e ^{3 t} )}{t ^{2}}

y = c_{1} e^{3 t} + c_{2} t e^{3 t} - e^{3 t} ln (t) - e^{3 t}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = \frac{( e ^{3 t} )}{t ^{2}}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} t e^{3 t} - e^{3 t} ln (t) - e^{3 t}

y = c_{1} e^{3 t} + c_{2} t e^{3 t} - e^{3 t} ln (t) - e^{3 t}

f (x) = \frac{2 x ^{2} + 2 x - 4}{x ^{2} + x - 6}

\frac{d}{d x} (2 (e^{- x} - e^{- x} x))

x \to \infty lim (\frac{x}{28})

\frac{\partial}{\partial x} (5 x^{3} y^{6})

\frac{\partial}{\partial x} (2 y - x)

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