What is the solution for x^{''}+2x+5x=1,x^'(0)=0,x(0)=0 ?

The solution for x^{''}+2x+5x=1,x^'(0)=0,x(0)=0 is x=-1/7 cos(sqrt(7)t)+1/7

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x^{''}+2x+5x=1,x^'(0)=0,x(0)=0

x^{^{''}} + 2 x + 5 x = 1, x^{^{'}} (0) = 0, x (0) = 0

x = - \frac{1}{7} cos (7 t) + \frac{1}{7}

Solution steps

x^{''} (t) + 2 x + 5 x = 1

Solve linear ODE:

x = - \frac{1}{7} cos (7 t) + \frac{1}{7}

x = - \frac{1}{7} cos (7 t) + \frac{1}{7}

(d^{2} - 4 d + 4) y = x^{2} + e^{x} + cos (2 x)

y^{^{''}} + y^{^{'}} - 6 y = - e^{2 x} + 12 x^{2} + 2 x + 3

y^{^{''}} + 6 y^{^{'}} + 25 x = 0

y^{^{''}} - n^{2} y = \frac{2 cos ( πn ) x}{n}

y^{^{''}} + 8 y^{^{'}} - 2 y = \frac{5}{e ^{3 t}}

What is the solution for x^{''}+2x+5x=1,x^'(0)=0,x(0)=0 ?
The solution for x^{''}+2x+5x=1,x^'(0)=0,x(0)=0 is x=-1/7 cos(sqrt(7)t)+1/7