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

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

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

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

y = c_{1} e^{(2 + 3) x} + c_{2} e^{(2 - 3) x} + \frac{1}{4} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} e^{(2 + 3) x} + c_{2} e^{(2 - 3) x} + \frac{1}{4} cos (x)

y = c_{1} e^{(2 + 3) x} + c_{2} e^{(2 - 3) x} + \frac{1}{4} cos (x)

y^{^{''}} + 8 y^{^{'}} + 16 y = 2 sin (4 x)

y^{^{''}} + 4 y^{^{'}} + 4 = \frac{1}{x ^{2} e ^{2 x}}

y^{^{''}} - y = 4 e^{t} - 2 e^{- t}, y (0) = 1, y^{^{'}} (0) = 4

(D^{4} - 1) y = x

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

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