What is the solution for y^{''}-2y^'+y=25cos(2x) ?

The solution for y^{''}-2y^'+y=25cos(2x) is y=c_{1}e^x+c_{2}xe^x-4sin(2x)-3cos(2x)

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y^{''}-2y^'+y=25cos(2x)

y^{^{''}} - 2 y^{^{'}} + y = 25 cos (2 x)

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

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = 25 cos (2 x)

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} = a y + b

y^{^{''}} + 6 y^{^{'}} + 5 y = 10 cos (5 t)

y^{^{''}} - 4 y^{^{'}} - 12 y = sin (3 t)

\frac{d ^{2} y}{d x ^{2}} + 9 y = - 15 x + 7 + e^{x}

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

What is the solution for y^{''}-2y^'+y=25cos(2x) ?
The solution for y^{''}-2y^'+y=25cos(2x) is y=c_{1}e^x+c_{2}xe^x-4sin(2x)-3cos(2x)