What is the solution for y^{''}-5y^'+6y=0,y(0)=2,y^'(0)=3 ?

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

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y^{''}-5y^'+6y=0,y(0)=2,y^'(0)=3

y^{^{''}} - 5 y^{^{'}} + 6 y = 0, y (0) = 2, y^{^{'}} (0) = 3

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

Solution steps

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

Solve linear ODE:

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

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

4 \frac{d ^{2} x}{d t ^{2}} + 12 \frac{d x}{d t} + 9 x = 0

y^{^{''''}} + 2 y^{^{'''}} + 2 y^{^{''}} = 0

\frac{d ^{2} y}{d x ^{2}} - 25 y = 0

9 y^{^{''}} + 6 y^{^{'}} + 82 y = 0, y (0) = 2

y^{^{''}} - 9 y = 0, y (0) = 2, y^{^{'}} (0) = 4

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