What is the solution for y^{''}-5y^'+6y=0,y(0)=0,y^'(0)=3 ?

The solution for y^{''}-5y^'+6y=0,y(0)=0,y^'(0)=3 is y=3e^{3t}-3e^{2t}

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y^{''}-5y^'+6y=0,y(0)=0,y^'(0)=3

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y = 3 e^{3 t} - 3 e^{2 t}

Solution steps

y^{''} (t) - 5 y^{'} (t) + 6 y = 0

Solve linear ODE:

y = 3 e^{3 t} - 3 e^{2 t}

y = 3 e^{3 t} - 3 e^{2 t}

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

y^{^{''}} - 7 y = 0

(D^{5} - 15 D^{3} + 10 D^{2} + 60 D - 72) y = 0

y^{^{'''}} - 7 y^{^{''}} - 49 y^{^{'}} + 343 y = 0

\frac{d ^{2} y}{d x ^{2}} + 7 \frac{d y}{d x} + 12 y = 0

What is the solution for y^{''}-5y^'+6y=0,y(0)=0,y^'(0)=3 ?
The solution for y^{''}-5y^'+6y=0,y(0)=0,y^'(0)=3 is y=3e^{3t}-3e^{2t}