What is the solution for x^{''}-3x^'+2x=0,x(0)=4,x^'(0)=6 ?

The solution for x^{''}-3x^'+2x=0,x(0)=4,x^'(0)=6 is x=2e^{2t}+2e^t

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x^{''}-3x^'+2x=0,x(0)=4,x^'(0)=6

x^{^{''}} - 3 x^{^{'}} + 2 x = 0, x (0) = 4, x^{^{'}} (0) = 6

x = 2 e^{2 t} + 2 e^{t}

Solution steps

x^{''} (t) - 3 x^{'} (t) + 2 x = 0

Solve linear ODE:

x = 2 e^{2 t} + 2 e^{t}

x = 2 e^{2 t} + 2 e^{t}

(D^{2} + D - 6) y = 0

y^{^{''}} + 8 y^{^{'}} - 9 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y (0) = - 2, y^{^{'}} (0) = - 8

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

θ^{^{''}} (t) + 49 θ (t) = 0

What is the solution for x^{''}-3x^'+2x=0,x(0)=4,x^'(0)=6 ?
The solution for x^{''}-3x^'+2x=0,x(0)=4,x^'(0)=6 is x=2e^{2t}+2e^t