What is the solution for y^{''}-12y^'+36y=0,y(1/6)=2,y(0)=3 ?

The solution for y^{''}-12y^'+36y=0,y(1/6)=2,y(0)=3 is y=3e^{6t}-(18e-12)e^{6t-1}t

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y^{''}-12y^'+36y=0,y(1/6)=2,y(0)=3

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (\frac{1}{6}) = 2, y (0) = 3

y = 3 e^{6 t} - (18 e - 12) e^{6 t - 1} t

Solution steps

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

Solve linear ODE:

y = 3 e^{6 t} - (18 e - 12) e^{6 t - 1} t

y = 3 e^{6 t} - (18 e - 12) e^{6 t - 1} t

16 x^{^{''}} + 40 x^{^{'}} + 169 x = 0

y^{^{''}} + 5 y^{^{'}} + 16 y = 0

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

10 x^{^{''}} + 30 x^{^{''}} + 20 x = 0, x (0) = - 7, x^{^{'}} (0) = - 2

y^{^{'''}} + y^{^{''}} - 22 y^{^{'}} - 40 y = 0

What is the solution for y^{''}-12y^'+36y=0,y(1/6)=2,y(0)=3 ?
The solution for y^{''}-12y^'+36y=0,y(1/6)=2,y(0)=3 is y=3e^{6t}-(18e-12)e^{6t-1}t