What is the solution for x^{''}-5x^'+6x=12,x(0)=2,x^'(0)=0 ?

The solution for x^{''}-5x^'+6x=12,x(0)=2,x^'(0)=0 is x=2

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x^{''}-5x^'+6x=12,x(0)=2,x^'(0)=0

x^{^{''}} - 5 x^{^{'}} + 6 x = 12, x (0) = 2, x^{^{'}} (0) = 0

x = 2

Solution steps

x^{''} (t) - 5 x^{'} (t) + 6 x = 12

Solve linear ODE:

x = 2

x = 2

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{x}

y^{^{''}} - 2 y^{^{'}} + y = e^{t - 1}, y (1) = 0, y^{^{'}} (1) = 5

y^{^{''}} - y^{^{'}} = x e^{2 x}

y^{^{''}} + 4 y = - 8 sin (2 t)

y^{^{''}} - 2 y^{^{'}} + 2 y = e^{2} x (cos (x) - 9 sin (x))

What is the solution for x^{''}-5x^'+6x=12,x(0)=2,x^'(0)=0 ?
The solution for x^{''}-5x^'+6x=12,x(0)=2,x^'(0)=0 is x=2