What is the solution for y^{''}-12y^'+36y=0,y(0)=5,y^'(0)=0 ?

The solution for y^{''}-12y^'+36y=0,y(0)=5,y^'(0)=0 is y=5e^{6t}-30e^{6t}t

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y^{''}-12y^'+36y=0,y(0)=5,y^'(0)=0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y = 5 e^{6 t} - 30 e^{6 t} t

Solution steps

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

Solve linear ODE:

y = 5 e^{6 t} - 30 e^{6 t} t

y = 5 e^{6 t} - 30 e^{6 t} t

y^{^{''}} + 3 y = 4 y^{^{'}}

y^{^{'''}} + 5 y^{^{''}} - 34 y^{^{'}} - 80 y = 0

y^{^{''}} + 1 y = 0, y (\frac{π}{2}) = 3, y^{^{'}} (\frac{π}{2}) = - 1

- 2 y^{^{''}} - 10 y^{^{'}} + \frac{5}{2} y = 0

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

What is the solution for y^{''}-12y^'+36y=0,y(0)=5,y^'(0)=0 ?
The solution for y^{''}-12y^'+36y=0,y(0)=5,y^'(0)=0 is y=5e^{6t}-30e^{6t}t