What is the solution for x^{''}-3x^'+2x=0,x^0=1,x^'0=-1 ?

The solution for x^{''}-3x^'+2x=0,x^0=1,x^'0=-1 is x=c_{1}e^{2t}+c_{2}e^t

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x^{''}-3x^'+2x=0,x^0=1,x^'0=-1

x^{^{''}} - 3 x^{^{'}} + 2 x = 0, x^{0} (0) = 1, x^{^{'}} 0 = - 1

x = c_{1} e^{2 t} + c_{2} e^{t}

Solution steps

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

Solve linear ODE:

x = c_{1} e^{2 t} + c_{2} e^{t}

x = c_{1} e^{2 t} + c_{2} e^{t}

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

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

What is the solution for x^{''}-3x^'+2x=0,x^0=1,x^'0=-1 ?
The solution for x^{''}-3x^'+2x=0,x^0=1,x^'0=-1 is x=c_{1}e^{2t}+c_{2}e^t