What is the solution for y^{''}+y=4,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+y=4,y(0)=0,y^'(0)=0 is y=-4cos(t)+4

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y^{''}+y=4,y(0)=0,y^'(0)=0

y^{^{''}} + y = 4, y (0) = 0, y^{^{'}} (0) = 0

y = - 4 cos (t) + 4

Solution steps

y^{''} (t) + y = 4

Solve linear ODE:

y = - 4 cos (t) + 4

y = - 4 cos (t) + 4

- 3 y^{^{'''}} + y^{^{''''}} = 27 c e^{3 x}

4 q^{^{''}} + 8 q^{^{'}} + \frac{1}{0.0125} q = 88 t

x, d (x) = \frac{1}{2} g x^{2}

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

\frac{d ^{2} y}{d x ^{2}} - 16 y = 9.6 e^{4 x} + 30 e^{x}

What is the solution for y^{''}+y=4,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+y=4,y(0)=0,y^'(0)=0 is y=-4cos(t)+4