What is the solution for y^{''}-12y^'+36y=0,y(0)=6,y^'(0)=1 ?

The solution for y^{''}-12y^'+36y=0,y(0)=6,y^'(0)=1 is y=6e^{6t}-35e^{6t}t

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y^{''}-12y^'+36y=0,y(0)=6,y^'(0)=1

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 6, y^{^{'}} (0) = 1

y = 6 e^{6 t} - 35 e^{6 t} t

Solution steps

y^{''} (t) - 12 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = 6 e^{6 t} - 35 e^{6 t} t

y = 6 e^{6 t} - 35 e^{6 t} t

y^{^{''}} + 6 y^{^{'}} + 35 y = 0

(4 D^{3} - 15 D^{2} + 5 D + 6) y = 0

y^{^{''''}} - k^{2} y = 0

y^{^{''}} + 5 y^{^{'}} - 24 y = 0, y (0) = a, y^{^{'}} (0) = 80

44 y^{^{'''}} + 15 y^{^{''}} + y^{^{'}} = 0

What is the solution for y^{''}-12y^'+36y=0,y(0)=6,y^'(0)=1 ?
The solution for y^{''}-12y^'+36y=0,y(0)=6,y^'(0)=1 is y=6e^{6t}-35e^{6t}t