What is the solution for y^{''}-2y^'+5y=8sin(x)-4cos(x) ?

The solution for y^{''}-2y^'+5y=8sin(x)-4cos(x) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+2sin(x)

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y^{''}-2y^'+5y=8sin(x)-4cos(x)

y^{^{''}} - 2 y^{^{'}} + 5 y = 8 sin (x) - 4 cos (x)

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + 5 y = 8 sin (x) - 4 cos (x)

Solve linear ODE:

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

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

y^{^{''}} + 9 y = sin^{4} (x)

\frac{d ^{2} x}{d t ^{2}} + 4 \frac{d x}{d t} + 4 x = 7 - 2 t^{2}

y^{^{''}} + 4 y = sin (3 t)

y^{^{''''}} - y = 3 t + cos (t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 1 + e^{x}

What is the solution for y^{''}-2y^'+5y=8sin(x)-4cos(x) ?
The solution for y^{''}-2y^'+5y=8sin(x)-4cos(x) is y=e^x(c_{1}cos(2x)+c_{2}sin(2x))+2sin(x)