What is the solution for y^{''}-4y^'+4y=-80sin(3x)-23cos(3x) ?

The solution for y^{''}-4y^'+4y=-80sin(3x)-23cos(3x) is y=c_{1}e^{2x}+c_{2}xe^{2x}+4sin(3x)-5cos(3x)

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y^{''}-4y^'+4y=-80sin(3x)-23cos(3x)

y^{^{''}} - 4 y^{^{'}} + 4 y = - 80 sin (3 x) - 23 cos (3 x)

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 4 sin (3 x) - 5 cos (3 x)

Solution steps

y^{''} (x) - 4 y^{'} (x) + 4 y = - 80 sin (3 x) - 23 cos (3 x)

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 4 sin (3 x) - 5 cos (3 x)

y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 4 sin (3 x) - 5 cos (3 x)

\frac{d ^{2}}{d t ^{2}} (θ (t)) + 4 \cdot θ (t) = csc^{2} (2 t)

y^{^{''}} - 4 y = x^{2}

y^{^{'''}} + y^{^{''}} = 12 x^{5}

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{- 3} e^{3 x}

y^{^{''}} + 3 y^{^{'}} + 2 y = 12 x^{2}

What is the solution for y^{''}-4y^'+4y=-80sin(3x)-23cos(3x) ?
The solution for y^{''}-4y^'+4y=-80sin(3x)-23cos(3x) is y=c_{1}e^{2x}+c_{2}xe^{2x}+4sin(3x)-5cos(3x)