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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

4 y^{^{''}} + y^{^{'}} + 6 y = 0

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

y^{^{''}} + 3 y^{^{'}} + 5 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''''}} + 3 y^{^{'''}} + y^{^{''}} = 0

2 y^{^{''}} + (a^{2} - 1) y = 0

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