What is the solution for y^{''}-4y^'+3y=20cos(x) ?

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

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

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

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

Solution steps

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

x^{^{''}} + x^{^{'}} + 12 x = 20 cos (3 t)

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

y^{^{'''}} - y^{^{'}} = x^{2} + 2 x e^{x} + e^{2 x}

y^{^{''}} + 6 y^{^{'}} + 8 y = 3 e^{5 x} + x^{2} - 1

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

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