What is the solution for y^{''}-y=4cos(t),y(0)=0,y^'(0)=1 ?

The solution for y^{''}-y=4cos(t),y(0)=0,y^'(0)=1 is y=(-1/2+2)e^t+1/2 e^{-t}-2cos(t)

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y^{''}-y=4cos(t),y(0)=0,y^'(0)=1

y^{^{''}} - y = 4 cos (t), y (0) = 0, y^{^{'}} (0) = 1

y = (- \frac{1}{2} + 2) e^{t} + \frac{1}{2} e^{- t} - 2 cos (t)

Solution steps

y^{''} (t) - y = 4 cos (t)

Solve linear ODE:

y = (- \frac{1}{2} + 2) e^{t} + \frac{1}{2} e^{- t} - 2 cos (t)

y = (- \frac{1}{2} + 2) e^{t} + \frac{1}{2} e^{- t} - 2 cos (t)

y^{^{''}} - 3 y^{^{'}} + 2 y = a, y (0) = 2, y^{^{'}} (0) = - 6

d (x) = \frac{2500}{30 ( 0.3 + 0.03 )}

y^{^{''}} = 2 y - x

y^{^{''}} - y \cdot a + g = 0

y^{^{'''}} - 11 y^{^{''}} + 28 y^{^{'}} = 36 e^{x}, y^{^{'}} (0) = 26

What is the solution for y^{''}-y=4cos(t),y(0)=0,y^'(0)=1 ?
The solution for y^{''}-y=4cos(t),y(0)=0,y^'(0)=1 is y=(-1/2+2)e^t+1/2 e^{-t}-2cos(t)