What is the solution for y^{''}-6y^'+10y=180sin(2x) ?

The solution for y^{''}-6y^'+10y=180sin(2x) is y=e^{3x}(c_{1}cos(x)+c_{2}sin(x))+6sin(2x)+12cos(2x)

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y^{''}-6y^'+10y=180sin(2x)

y^{^{''}} - 6 y^{^{'}} + 10 y = 180 sin (2 x)

y = e^{3 x} (c_{1} cos (x) + c_{2} sin (x)) + 6 sin (2 x) + 12 cos (2 x)

Solution steps

y^{''} (x) - 6 y^{'} (x) + 10 y = 180 sin (2 x)

Solve linear ODE:

y = e^{3 x} (c_{1} cos (x) + c_{2} sin (x)) + 6 sin (2 x) + 12 cos (2 x)

y = e^{3 x} (c_{1} cos (x) + c_{2} sin (x)) + 6 sin (2 x) + 12 cos (2 x)

y^{^{''}} - 9 y^{^{'}} + 20 y = e^{x} (- 9 - 78 x + 36 x^{2})

y^{^{''}} - 3 y^{^{'}} - 10 y = - 18 t + 20 t^{2}

y^{^{'''}} - y^{^{'}} = e^{t} (1 + t)

(d^{2} - 2 d + 5) y = e^{2 x} \cdot sin (x)

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = x^{3} + 2 x^{2} + 5 x + 2

What is the solution for y^{''}-6y^'+10y=180sin(2x) ?
The solution for y^{''}-6y^'+10y=180sin(2x) is y=e^{3x}(c_{1}cos(x)+c_{2}sin(x))+6sin(2x)+12cos(2x)