What is the solution for (d^4-1)y=e^xcos(x)cos^3(x) ?

The solution for (d^4-1)y=e^xcos(x)cos^3(x) is

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(d^4-1)y=e^xcos(x)cos^3(x)

(d^{4} - 1) y = e^{x} cos (x) cos^{3} (x)

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} cos (x) + c_{4} sin (x) + \frac{1}{4} e^{x} (cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x))) + e^{- x} \cdot \int - \frac{e ^{2 x} cos ^{4} ( x )}{4} d x + cos (x) \cdot \int \frac{e ^{x} cos ^{4} ( x ) sin ( x )}{2} d x + sin (x) \cdot \int - \frac{e ^{x} cos ^{5} ( x )}{2} d x

Solution steps

(d^{4} - 1) y = e^{x} cos (x) cos^{3} (x)

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} cos (x) + c_{4} sin (x) + \frac{1}{4} e^{x} (cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x))) + e^{- x} \cdot \int - \frac{e ^{2 x} cos ^{4} ( x )}{4} d x + cos (x) \cdot \int \frac{e ^{x} cos ^{4} ( x ) sin ( x )}{2} d x + sin (x) \cdot \int - \frac{e ^{x} cos ^{5} ( x )}{2} d x

y = c_{1} e^{x} + c_{2} e^{- x} + c_{3} cos (x) + c_{4} sin (x) + \frac{1}{4} e^{x} (cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x))) + e^{- x} \cdot \int - \frac{e ^{2 x} cos ^{4} ( x )}{4} d x + cos (x) \cdot \int \frac{e ^{x} cos ^{4} ( x ) sin ( x )}{2} d x + sin (x) \cdot \int - \frac{e ^{x} cos ^{5} ( x )}{2} d x

(d^{2} - 4 d + 4) \cdot y = 8 e^{2 x} \cdot x^{2} \cdot sin^{2} (x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 t - 8

(d^{3} - 2 d + 4) y = 3 x^{2} - 5 x + 2

(d^{3} - 3 d^{2} + 4 d - 2) \cdot y = e^{x} + cos (x) + x

(d^{4} + d^{2} + 16) y = 16 x^{2} + 256

What is the solution for (d^4-1)y=e^xcos(x)cos^3(x) ?
The solution for (d^4-1)y=e^xcos(x)cos^3(x) is