What is the solution for 12y^{''}-13y^'+y=0,y(0)=12,y^'(0)= 1/12 ?

The solution for 12y^{''}-13y^'+y=0,y(0)=12,y^'(0)= 1/12 is y=-e^t+13e^{t/(12)}

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12y^{''}-13y^'+y=0,y(0)=12,y^'(0)= 1/12

12 y^{^{''}} - 13 y^{^{'}} + y = 0, y (0) = 12, y^{^{'}} (0) = \frac{1}{12}

y = - e^{t} + 13 e^{\frac{t}{12}}

Solution steps

12 y^{''} (t) - 13 y^{'} (t) + y = 0

Solve linear ODE:

y = - e^{t} + 13 e^{\frac{t}{12}}

y = - e^{t} + 13 e^{\frac{t}{12}}

16 y^{^{''}} + 12 y^{^{'}} + 9 y = 0

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

(D^{2} - 9 D + 18) y = 0

16 y^{^{''}} = - 49 y^{^{'}}

y^{^{''''}} + 16 y^{^{''}} = 0

What is the solution for 12y^{''}-13y^'+y=0,y(0)=12,y^'(0)= 1/12 ?
The solution for 12y^{''}-13y^'+y=0,y(0)=12,y^'(0)= 1/12 is y=-e^t+13e^{t/(12)}