What is the solution for y^{''}+3y^'-4y=2cos(3x) ?

The solution for y^{''}+3y^'-4y=2cos(3x) is y=c_{1}e^x+c_{2}e^{-4x}+9/125 sin(3x)-13/125 cos(3x)

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

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

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{9}{125} sin (3 x) - \frac{13}{125} cos (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{9}{125} sin (3 x) - \frac{13}{125} cos (3 x)

y = c_{1} e^{x} + c_{2} e^{- 4 x} + \frac{9}{125} sin (3 x) - \frac{13}{125} cos (3 x)

y^{^{''}} + 2 y^{^{'}} + y = t^{2} - 6

y^{^{''}} + 2 y^{^{'}} + 2 y = \frac{e ^{- x}}{cos ( x )}

y^{^{''}} - y = ln (t)

y^{^{''}} - 12 y^{^{'}} + 36 y = 12 x + 5

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

What is the solution for y^{''}+3y^'-4y=2cos(3x) ?
The solution for y^{''}+3y^'-4y=2cos(3x) is y=c_{1}e^x+c_{2}e^{-4x}+9/125 sin(3x)-13/125 cos(3x)