What is the solution for y^{''}-2y^'+y=cos(3x) ?

The solution for y^{''}-2y^'+y=cos(3x) is y=c_{1}e^x+c_{2}xe^x-3/50 sin(3x)-2/25 cos(3x)

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

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

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{3}{50} sin (3 x) - \frac{2}{25} cos (3 x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{3}{50} sin (3 x) - \frac{2}{25} cos (3 x)

y = c_{1} e^{x} + c_{2} x e^{x} - \frac{3}{50} sin (3 x) - \frac{2}{25} cos (3 x)

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

y^{^{''}} - 4 y = e^{x} cos (x), y (0) = 1, y^{^{'}} (0) = 2

y^{^{'''}} + 6 y^{^{''}} + 12 y^{^{'}} + 8 y = 5 e^{- 2 x}

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} + 10 y = e^{2 x} + 20

y^{^{''}} - 2 y^{^{'}} - 35 y = - 97 t + 70 t^{2}

What is the solution for y^{''}-2y^'+y=cos(3x) ?
The solution for y^{''}-2y^'+y=cos(3x) is y=c_{1}e^x+c_{2}xe^x-3/50 sin(3x)-2/25 cos(3x)