What is the solution for (d^4-1)y=e^xcos(x) ?

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

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(d^4-1)y=e^xcos(x)

(d^{4} - 1) y = e^{x} cos (x)

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

Solution steps

(d^{4} - 1) y = e^{x} cos (x)

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} = x^{2} - x

(d^{2} + 9) y = 7 e^{3 x} + 3 x^{2} + 4

y^{^{''}} - 2 y^{^{'}} + y = (3 x + 1) e^{x}

(d^{2} - 13 d + 12) \cdot y = x

(d - 1)^{2} \cdot y = e^{x} \cdot sec^{2} (x) tan (x)

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