What is the solution for y^{''}+4y=4,y(0)=1,y^'(0)=0 ?

The solution for y^{''}+4y=4,y(0)=1,y^'(0)=0 is y=1

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y^{''}+4y=4,y(0)=1,y^'(0)=0

y^{^{''}} + 4 y = 4, y (0) = 1, y^{^{'}} (0) = 0

y = 1

Solution steps

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

Solve linear ODE:

y = 1

y = 1

y^{^{''}} + 3 y^{^{'}} + 2 y = u_{2} (t), y (0) = 6, y^{^{'}} (0) = 8

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

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

y^{^{''}} - 9 y = sin (t) - cos (t)

\frac{d ^{2} y}{d x ^{2}} - \frac{d y}{d x} - 2 y = 6 x + 6 e^{- x}

What is the solution for y^{''}+4y=4,y(0)=1,y^'(0)=0 ?
The solution for y^{''}+4y=4,y(0)=1,y^'(0)=0 is y=1