What is the solution for y^{''}+2y+y=e^{-t}ln(t) ?

The solution for y^{''}+2y+y=e^{-t}ln(t) is

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y^{''}+2y+y=e^{-t}ln(t)

y^{^{''}} + 2 y + y = e^{- t} ln (t)

y = c_{1} cos (3 t) + c_{2} sin (3 t) + cos (3 t) \cdot \int - \frac{3 e ^{- t} sin ( 3 t ) ln ( t )}{3} d t + sin (3 t) \cdot \int \frac{3 e ^{- t} cos ( 3 t ) ln ( t )}{3} d t

Solution steps

y^{''} (t) + 2 y + y = e^{- t} ln (t)

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + cos (3 t) \cdot \int - \frac{3 e ^{- t} sin ( 3 t ) ln ( t )}{3} d t + sin (3 t) \cdot \int \frac{3 e ^{- t} cos ( 3 t ) ln ( t )}{3} d t

y = c_{1} cos (3 t) + c_{2} sin (3 t) + cos (3 t) \cdot \int - \frac{3 e ^{- t} sin ( 3 t ) ln ( t )}{3} d t + sin (3 t) \cdot \int \frac{3 e ^{- t} cos ( 3 t ) ln ( t )}{3} d t

y^{^{''}} - 10 y^{^{'}} + 25 y = t, y (0) = 0, y^{^{'}} (0) = 1

y^{^{'''}} - 49 y^{^{'}} = t e^{- 7 t} + 8 cos (7 t)

y^{^{''}} - y^{^{'}} - 2 y = e^{- t}

y^{^{''''}} + 2 y^{^{''}} + y = 3 + cos (6 t)

y^{^{''}} + 4 y^{^{'}} = 3 sin (2 x)

What is the solution for y^{''}+2y+y=e^{-t}ln(t) ?
The solution for y^{''}+2y+y=e^{-t}ln(t) is