What is the solution for y^{''}-10y^'+25y=t^{-2}e^{5t} ?

The solution for y^{''}-10y^'+25y=t^{-2}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}-e^{5t}ln(t)-e^{5t}

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y^{''}-10y^'+25y=t^{-2}e^{5t}

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 2} e^{5 t}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} - e^{5 t} ln (t) - e^{5 t}

Solution steps

y^{''} (t) - 10 y^{'} (t) + 25 y = t^{- 2} e^{5 t}

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} t e^{5 t} - e^{5 t} ln (t) - e^{5 t}

y = c_{1} e^{5 t} + c_{2} t e^{5 t} - e^{5 t} ln (t) - e^{5 t}

\frac{d}{d t} (sin^{3} (t))

x (\frac{d y}{d x}) + (3 x + 1) y = e^{- 3 x}

\int_{0}^{4} 3 d x

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

\int_{11}^{12} 14 arccot (x) d x

What is the solution for y^{''}-10y^'+25y=t^{-2}e^{5t} ?
The solution for y^{''}-10y^'+25y=t^{-2}e^{5t} is y=c_{1}e^{5t}+c_{2}te^{5t}-e^{5t}ln(t)-e^{5t}