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

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

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

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

x = e^{- t} (- cos (t) - sin (t)) + 1

Solution steps

x^{''} (t) + 2 x^{'} (t) + 2 x = 2

Solve linear ODE:

x = e^{- t} (- cos (t) - sin (t)) + 1

x = e^{- t} (- cos (t) - sin (t)) + 1

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

y^{^{''}} = 2 y + tan^{3} (x), y^{1} (x) = \frac{1}{2} tan (x)

y^{^{''}} - 2 y + y = \frac{e ^{x}}{1 + x ^{2}}

y^{^{''}} - 11 y^{^{'}} + 28 y = 108 e^{t}

y^{^{''}} - 5 y^{^{'}} + 6 y = sin (3 x)

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