What is the solution for y^{''}-7y^'+12y=0,y(0)=2,y^'(0)=3 ?

The solution for y^{''}-7y^'+12y=0,y(0)=2,y^'(0)=3 is y=-3e^{4t}+5e^{3t}

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y^{''}-7y^'+12y=0,y(0)=2,y^'(0)=3

y^{^{''}} - 7 y^{^{'}} + 12 y = 0, y (0) = 2, y^{^{'}} (0) = 3

y = - 3 e^{4 t} + 5 e^{3 t}

Solution steps

y^{''} (t) - 7 y^{'} (t) + 12 y = 0

Solve linear ODE:

y = - 3 e^{4 t} + 5 e^{3 t}

y = - 3 e^{4 t} + 5 e^{3 t}

y^{^{''''}} - y^{^{'''}} - 3 y^{^{''}} + 17 y^{^{'}} - 30 y = 0

(d^{3} - d^{2} - 4 d + 4) y = 0

4 y^{^{''}} + \frac{1}{784} y = 0, y (0) = 7

64 y^{^{''}} + 144 y^{^{'}} + 83 y = 0

x^{^{''}} + 2 \cdot x^{^{'}} + x = 0

What is the solution for y^{''}-7y^'+12y=0,y(0)=2,y^'(0)=3 ?
The solution for y^{''}-7y^'+12y=0,y(0)=2,y^'(0)=3 is y=-3e^{4t}+5e^{3t}