What is the solution for y^{''}+2y^'-3y=-9cos(2x)-33sin(2x) ?

The solution for y^{''}+2y^'-3y=-9cos(2x)-33sin(2x) is y=c_{1}e^x+c_{2}e^{-3x}+3sin(2x)+3cos(2x)

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y^{''}+2y^'-3y=-9cos(2x)-33sin(2x)

y^{^{''}} + 2 y^{^{'}} - 3 y = - 9 cos (2 x) - 33 sin (2 x)

y = c_{1} e^{x} + c_{2} e^{- 3 x} + 3 sin (2 x) + 3 cos (2 x)

Solution steps

y^{''} (x) + 2 y^{'} (x) - 3 y = - 9 cos (2 x) - 33 sin (2 x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} + 3 sin (2 x) + 3 cos (2 x)

y = c_{1} e^{x} + c_{2} e^{- 3 x} + 3 sin (2 x) + 3 cos (2 x)

y^{^{''}} - 8 y^{^{'}} + 16 y = 12 x + 6

(d^{2} + 4) y = cos (2 x) + cos (4 x)

y^{^{'''}} - y^{^{''}} - 25 y^{^{'}} + 25 y = 9 - e^{x} + e^{5 x}

y^{^{''}} + 4 y^{^{'}} + 13 y = t e^{- t}, y (0) = 0, y^{^{'}} (0) = 4

D (T) = - 2

What is the solution for y^{''}+2y^'-3y=-9cos(2x)-33sin(2x) ?
The solution for y^{''}+2y^'-3y=-9cos(2x)-33sin(2x) is y=c_{1}e^x+c_{2}e^{-3x}+3sin(2x)+3cos(2x)