What is the solution for y^{''}-6y^'+9y=0,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-6y^'+9y=0,y(0)=0,y^'(0)=1 is y=e^{3t}t

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y^{''}-6y^'+9y=0,y(0)=0,y^'(0)=1

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = e^{3 t} t

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{3 t} t

y = e^{3 t} t

c^{2} \cdot f^{^{''}} (t) = a f^{^{''}} (t) - b f (t)

(y^{^{''}} + 4 y^{^{'}} + 20 y)^{2} = 0

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

y^{^{'''}} - 5 y^{^{''}} - 2 y^{^{'}} + 24 y = 0

f^{^{''}} (x) + f (x) = 0, f (0) = 1

What is the solution for y^{''}-6y^'+9y=0,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-6y^'+9y=0,y(0)=0,y^'(0)=1 is y=e^{3t}t