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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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