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

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

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

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

x = cos (t) + sin (t) + 1

Solution steps

x^{''} (t) + x = 1

Solve linear ODE:

x = cos (t) + sin (t) + 1

x = cos (t) + sin (t) + 1

y^{^{''}} + 9 y = 12 cos (3 x)

2 y^{^{''}} + 3 y^{^{'}} + y = x^{2}

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

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

y^{^{''}} + y = 6 e^{- 7 x}, y (0) = 0, y^{^{'}} (0) = 0

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