What is the solution for y^{''}+4y=4cos(x)+3sin(x)+8 ?

The solution for y^{''}+4y=4cos(x)+3sin(x)+8 is y=c_{1}cos(2x)+c_{2}sin(2x)+4/3 cos(x)+sin(x)+2

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y^{''}+4y=4cos(x)+3sin(x)+8

y^{^{''}} + 4 y = 4 cos (x) + 3 sin (x) + 8

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{4}{3} cos (x) + sin (x) + 2

Solution steps

y^{''} (x) + 4 y = 4 cos (x) + 3 sin (x) + 8

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{4}{3} cos (x) + sin (x) + 2

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{4}{3} cos (x) + sin (x) + 2

y^{^{''}} + 4 y^{^{'}} + 20 y = - 3 sin (2 x)

y^{^{'''}} - y^{^{'}} = 11 sin (t)

(D^{2} - D - 2) y = 18 x e^{- x}

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

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 3 e^{- t} + 3

What is the solution for y^{''}+4y=4cos(x)+3sin(x)+8 ?
The solution for y^{''}+4y=4cos(x)+3sin(x)+8 is y=c_{1}cos(2x)+c_{2}sin(2x)+4/3 cos(x)+sin(x)+2