What is the solution for y^{''}-4y^'+3y=13cos(2x) ?

The solution for y^{''}-4y^'+3y=13cos(2x) is y=c_{1}e^{3x}+c_{2}e^x-8/5 sin(2x)-1/5 cos(2x)

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y^{''}-4y^'+3y=13cos(2x)

y^{^{''}} - 4 y^{^{'}} + 3 y = 13 cos (2 x)

y = c_{1} e^{3 x} + c_{2} e^{x} - \frac{8}{5} sin (2 x) - \frac{1}{5} cos (2 x)

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = 13 cos (2 x)

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} - \frac{8}{5} sin (2 x) - \frac{1}{5} cos (2 x)

y = c_{1} e^{3 x} + c_{2} e^{x} - \frac{8}{5} sin (2 x) - \frac{1}{5} cos (2 x)

y^{^{'''}} - 6 y^{^{''}} + 12 y^{^{'}} - 8 y = t e^{3} + 1

y^{^{''}} + y^{^{'}} - 2 y = 2 (3 x + 1) e^{x} + 20 cos (2 x)

y^{^{''}} + 12 y = 8 e^{4 x}

y^{^{''}} + 2 y^{^{'}} + y = \frac{t}{e ^{t}}

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

What is the solution for y^{''}-4y^'+3y=13cos(2x) ?
The solution for y^{''}-4y^'+3y=13cos(2x) is y=c_{1}e^{3x}+c_{2}e^x-8/5 sin(2x)-1/5 cos(2x)