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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{'''}} - y^{^{''}} - y^{^{'}} + y = 7 e^{- t} + 4

y^{^{''}} + y^{^{'}} + 2 y = sin (t)

\frac{d ^{2} x}{d t ^{2}} + 6 \cdot \frac{d x}{d t} + 8 x = 2 e^{- 3 t}

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

y^{^{''}} + 6 y^{^{'}} + 9 y = 50 e^{2 x}

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