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

The solution for y^{''}+4y=sin(x)-cos(x) is y=c_{1}cos(2x)+c_{2}sin(2x)+1/3 sin(x)-1/3 cos(x)

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

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

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

Solution steps

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

Solve linear ODE:

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

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

x^{^{''}} - 6 x^{^{'}} + 9 x = 2 t e^{3 t}

y^{^{''}} - 3 y^{^{'}} + y = - e^{x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{e ^{- x}}{1 - x ^{2}}

y^{^{''}} - 5 y^{^{'}} + 4 y = x, y (0) = 0, y^{^{'}} (0) = 0

\frac{d ^{2} y}{d t ^{2}} - 2 \frac{d y}{d t} - 35 y = (t + 1) e^{t}

What is the solution for y^{''}+4y=sin(x)-cos(x) ?
The solution for y^{''}+4y=sin(x)-cos(x) is y=c_{1}cos(2x)+c_{2}sin(2x)+1/3 sin(x)-1/3 cos(x)