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

The solution for y^{''}-6y^'+9y=0,y(0)=6,y^'(0)=0 is y=6e^{3t}-18e^{3t}t

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

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

y = 6 e^{3 t} - 18 e^{3 t} t

Solution steps

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

Solve linear ODE:

y = 6 e^{3 t} - 18 e^{3 t} t

y = 6 e^{3 t} - 18 e^{3 t} t

f (x) = \frac{sin ( x )}{2 + cos ( x )}

x \to 0 lim (x^{2} + 1)

\frac{d y}{d x} = x + 4

30 e^{x} x

\int_{0}^{1} (u + 2) (u - 3) d u

What is the solution for y^{''}-6y^'+9y=0,y(0)=6,y^'(0)=0 ?
The solution for y^{''}-6y^'+9y=0,y(0)=6,y^'(0)=0 is y=6e^{3t}-18e^{3t}t