What is the solution for 4y^{''}-12y^'+9y=0,y(0)=1 ?

The solution for 4y^{''}-12y^'+9y=0,y(0)=1 is y=e^{(3t)/2}+c_{2}e^{(3t)/2}t

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4y^{''}-12y^'+9y=0,y(0)=1

4 y^{^{''}} - 12 y^{^{'}} + 9 y = 0, y (0) = 1

y = e^{\frac{3 t}{2}} + c_{2} e^{\frac{3 t}{2}} t

Solution steps

4 y^{''} (t) - 12 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{2}} + c_{2} e^{\frac{3 t}{2}} t

y = e^{\frac{3 t}{2}} + c_{2} e^{\frac{3 t}{2}} t

(d^{3} + 9 d^{2} + 27 d + 27) y = 0

(D^{2} - D - 2) y = 0

y^{^{''}} + 5 y^{^{'}} - 36 y = 0, y (0) = α, y^{^{'}} (0) = - 72

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 1, y^{^{'}} (0) = 29

y^{^{''}} + (\frac{9.8}{0.06}) y = 0

What is the solution for 4y^{''}-12y^'+9y=0,y(0)=1 ?
The solution for 4y^{''}-12y^'+9y=0,y(0)=1 is y=e^{(3t)/2}+c_{2}e^{(3t)/2}t