What is the solution for (D^2-4D+3)y=20cos(x) ?

The solution for (D^2-4D+3)y=20cos(x) is y=c_{1}e^{3x}+c_{2}e^x-4sin(x)+2cos(x)

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(D^2-4D+3)y=20cos(x)

(D^{2} - 4 D + 3) y = 20 cos (x)

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

Solution steps

(D^{2} - 4 D + 3) y = 20 cos (x)

Solve linear ODE:

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

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

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

y^{^{''}} - 9 y = 20 sin (x) + 3 x^{2}

y^{^{''}} + y^{^{'}} + y = k x

y^{^{''}} - y^{^{'}} - 2 y = 2 x^{3} - 3 x^{2} + 8 x + 1

y^{^{''}} + 16 y = 8 e^{3 x}

What is the solution for (D^2-4D+3)y=20cos(x) ?
The solution for (D^2-4D+3)y=20cos(x) is y=c_{1}e^{3x}+c_{2}e^x-4sin(x)+2cos(x)