What is the solution for y^{''}-6y^'+9y=t,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-6y^'+9y=t,y(0)=0,y^'(0)=0 is y=-2/27 e^{3t}+(e^{3t}t)/9+t/9+2/27

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y^{''}-6y^'+9y=t,y(0)=0,y^'(0)=0

y^{^{''}} - 6 y^{^{'}} + 9 y = t, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{2}{27} e^{3 t} + \frac{e ^{3 t} t}{9} + \frac{t}{9} + \frac{2}{27}

Solution steps

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

Solve linear ODE:

y = - \frac{2}{27} e^{3 t} + \frac{e ^{3 t} t}{9} + \frac{t}{9} + \frac{2}{27}

y = - \frac{2}{27} e^{3 t} + \frac{e ^{3 t} t}{9} + \frac{t}{9} + \frac{2}{27}

y^{^{''}} - y = x e^{\frac{x}{2}}

y^{^{''}} + 9 y = e^{2 x} + x^{2} sin (x)

y^{^{''}} - 2 y^{^{'}} + y = 4 x^{2} - 4 + x^{- 1} e^{x}

y^{^{''}} + y = x e^{x} cos (2 x)

y^{^{''}} + 6 y^{^{'}} + 8 y = 2 t + e^{t}

What is the solution for y^{''}-6y^'+9y=t,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-6y^'+9y=t,y(0)=0,y^'(0)=0 is y=-2/27 e^{3t}+(e^{3t}t)/9+t/9+2/27