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

The solution for xy^{'''}-4xy^'+2xy=cos(x) is

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

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

y = c_{1} e^{0.53918 \dots x} + c_{2} e^{1.67513 \dots x} + c_{3} e^{- 2.21431 \dots x} + e^{0.53918 \dots x} \cdot \int - \frac{3.88945 \dots e ^{- 0.53918 \dots x} cos ( x )}{12.16552 \dots x} d x + e^{1.67513 \dots x} \cdot \int \frac{2.75350 \dots e ^{- 1.67513 \dots x} cos ( x )}{12.16552 \dots x} d x + e^{- 2.21431 \dots x} \cdot \int \frac{1.13594 \dots e ^{2.21431 \dots x} cos ( x )}{12.16552 \dots x} d x

Solution steps

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

Solve linear ODE:

y = c_{1} e^{0.53918 \dots x} + c_{2} e^{1.67513 \dots x} + c_{3} e^{- 2.21431 \dots x} + e^{0.53918 \dots x} \cdot \int - \frac{3.88945 \dots e ^{- 0.53918 \dots x} cos ( x )}{12.16552 \dots x} d x + e^{1.67513 \dots x} \cdot \int \frac{2.75350 \dots e ^{- 1.67513 \dots x} cos ( x )}{12.16552 \dots x} d x + e^{- 2.21431 \dots x} \cdot \int \frac{1.13594 \dots e ^{2.21431 \dots x} cos ( x )}{12.16552 \dots x} d x

y = c_{1} e^{0.53918 \dots x} + c_{2} e^{1.67513 \dots x} + c_{3} e^{- 2.21431 \dots x} + e^{0.53918 \dots x} \cdot \int - \frac{3.88945 \dots e ^{- 0.53918 \dots x} cos ( x )}{12.16552 \dots x} d x + e^{1.67513 \dots x} \cdot \int \frac{2.75350 \dots e ^{- 1.67513 \dots x} cos ( x )}{12.16552 \dots x} d x + e^{- 2.21431 \dots x} \cdot \int \frac{1.13594 \dots e ^{2.21431 \dots x} cos ( x )}{12.16552 \dots x} d x

x^{^{''}} - 4 x^{^{'}} + 3 x = t e^{t}

y^{^{''}} - 4 y^{^{'}} + 3 y = 13 cos (2 x)

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = t e^{3} + 1

y^{^{''}} + y^{^{'}} - 2 y = 2 (3 x + 1) e^{x} + 20 cos (2 x)

y^{^{''}} + 12 y = 8 e^{4 x}

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