What is the solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.455 ?

The solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.455 is y=4e^{6t}-8.545e^{6t}t

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y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.455

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.455

y = 4 e^{6 t} - 8.545 e^{6 t} t

Solution steps

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

Solve linear ODE:

y = 4 e^{6 t} - 8.545 e^{6 t} t

y = 4 e^{6 t} - 8.545 e^{6 t} t

3 y^{^{''}} + 4 y^{^{'}} + 5 y = 0

y^{^{''}} + \frac{49}{3} y = 0

y^{^{''}} - 5 y^{^{''}} + 24 y^{^{'}} - 20 y = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 0, y (0) = 4, y^{^{'}} (0) = 15.449

y^{^{''''}} + 8 y^{^{'''}} + 21 y^{^{''}} + 22 y^{^{'}} + 8 y = 0

What is the solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.455 ?
The solution for y^{''}-12y^'+36y=0,y(0)=4,y^'(0)=15.455 is y=4e^{6t}-8.545e^{6t}t