What is the solution for y^{''}+5y^'+4y=8x^2+3+2cos(2x) ?

The solution for y^{''}+5y^'+4y=8x^2+3+2cos(2x) is y=c_{1}e^{-x}+c_{2}e^{-4x}+2x^2-5x+6+1/5 sin(2x)

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y^{''}+5y^'+4y=8x^2+3+2cos(2x)

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + 3 + 2 cos (2 x)

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + 2 x^{2} - 5 x + 6 + \frac{1}{5} sin (2 x)

Solution steps

y^{''} (x) + 5 y^{'} (x) + 4 y = 8 x^{2} + 3 + 2 cos (2 x)

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + 2 x^{2} - 5 x + 6 + \frac{1}{5} sin (2 x)

y = c_{1} e^{- x} + c_{2} e^{- 4 x} + 2 x^{2} - 5 x + 6 + \frac{1}{5} sin (2 x)

What is the solution for y^{''}+5y^'+4y=8x^2+3+2cos(2x) ?
The solution for y^{''}+5y^'+4y=8x^2+3+2cos(2x) is y=c_{1}e^{-x}+c_{2}e^{-4x}+2x^2-5x+6+1/5 sin(2x)