What is the solution for y^{''}-5y^'+6y=t,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-5y^'+6y=t,y(0)=0,y^'(0)=1 is y=(5/4-5/36)e^{3t}-5/4 e^{2t}+t/6+5/36

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y^{''}-5y^'+6y=t,y(0)=0,y^'(0)=1

y^{^{''}} - 5 y^{^{'}} + 6 y = t, y (0) = 0, y^{^{'}} (0) = 1

y = (\frac{5}{4} - \frac{5}{36}) e^{3 t} - \frac{5}{4} e^{2 t} + \frac{t}{6} + \frac{5}{36}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 6 y = t

Solve linear ODE:

y = (\frac{5}{4} - \frac{5}{36}) e^{3 t} - \frac{5}{4} e^{2 t} + \frac{t}{6} + \frac{5}{36}

y = (\frac{5}{4} - \frac{5}{36}) e^{3 t} - \frac{5}{4} e^{2 t} + \frac{t}{6} + \frac{5}{36}

y^{^{'''}} + 18 y^{^{''}} + 108 y^{^{'}} + 216 y = e^{- 6 x}

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

9 y^{^{''}} + 2 y^{^{'}} - y = 15

y^{^{''}} - y^{^{'}} = 1, y (0) = 18, y^{^{'}} (0) = 1

y^{^{''}} + 5 y = (x)

What is the solution for y^{''}-5y^'+6y=t,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-5y^'+6y=t,y(0)=0,y^'(0)=1 is y=(5/4-5/36)e^{3t}-5/4 e^{2t}+t/6+5/36