What is the solution for y^{''}+4y=3,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+4y=3,y(0)=0,y^'(0)=0 is y=-3/4 cos(2t)+3/4

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y^{''}+4y=3,y(0)=0,y^'(0)=0

y^{^{''}} + 4 y = 3, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{3}{4} cos (2 t) + \frac{3}{4}

Solution steps

y^{''} (t) + 4 y = 3

Solve linear ODE:

y = - \frac{3}{4} cos (2 t) + \frac{3}{4}

y = - \frac{3}{4} cos (2 t) + \frac{3}{4}

y^{^{''}} + 4 y^{^{'}} + 4 = e^{- t}

y^{^{''}} - 4 y^{^{'}} + 4 y = 3

(D^{2} + 4) y = sinh (x) sin (2 x)

y^{^{''}} + 4 y^{^{'}} - 12 y = 3 x^{2} + x

2 y^{^{''}} + y^{^{'}} - 10 y = x e^{3 x} cos (π x)

What is the solution for y^{''}+4y=3,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=3,y(0)=0,y^'(0)=0 is y=-3/4 cos(2t)+3/4