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

The solution for y^{''}-3y^'-4y=sin(x) is y=c_{1}e^{4x}+c_{2}e^{-x}-5/34 sin(x)+3/34 cos(x)

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

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

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{5}{34} sin (x) + \frac{3}{34} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{5}{34} sin (x) + \frac{3}{34} cos (x)

y = c_{1} e^{4 x} + c_{2} e^{- x} - \frac{5}{34} sin (x) + \frac{3}{34} cos (x)

y^{^{''}} + 4 y^{^{'}} + 32 y = 4 cos (4 t)

y^{^{''}} + 4 y = t^{2} + 6 e^{t}, y (0) = 0, y^{^{'}} (0) = 7

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{x ^{2} + 4}

y^{^{''}} + 4 y = 3 sin (t), y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 25 y = \frac{2}{3} cos (2) t

What is the solution for y^{''}-3y^'-4y=sin(x) ?
The solution for y^{''}-3y^'-4y=sin(x) is y=c_{1}e^{4x}+c_{2}e^{-x}-5/34 sin(x)+3/34 cos(x)