What is the solution for y^{''}-y=2.9cos(t),y(0)=0,y^'(0)=2.9 ?

The solution for y^{''}-y=2.9cos(t),y(0)=0,y^'(0)=2.9 is y=2.175e^t-0.725e^{-t}-1.45cos(t)

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y^{''}-y=2.9cos(t),y(0)=0,y^'(0)=2.9

y^{^{''}} - y = 2.9 cos (t), y (0) = 0, y^{^{'}} (0) = 2.9

y = 2.175 e^{t} - 0.725 e^{- t} - 1.45 cos (t)

Solution steps

y^{''} (t) - y = 2.9 cos (t)

Solve linear ODE:

y = 2.175 e^{t} - 0.725 e^{- t} - 1.45 cos (t)

y = 2.175 e^{t} - 0.725 e^{- t} - 1.45 cos (t)

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\frac{d ^{2} y}{d t ^{2}} - 4 \frac{d y}{d t} + 4 y = e^{- 4 t}

y^{^{''}} + 8 y^{^{'}} + 15 y = - 6 cos (5 x)

x^{^{''}} + b x^{^{'}} - c = 0

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} = x^{2}

What is the solution for y^{''}-y=2.9cos(t),y(0)=0,y^'(0)=2.9 ?
The solution for y^{''}-y=2.9cos(t),y(0)=0,y^'(0)=2.9 is y=2.175e^t-0.725e^{-t}-1.45cos(t)