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

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

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

linear

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8} + \frac{5}{3} cos (x)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8} + \frac{5}{3} cos (x)

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x ^{2}}{4} - \frac{1}{8} + \frac{5}{3} cos (x)

y^{^{''}} - y^{^{'}} - 6 y = sin (x)

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

y^{^{''}} - 6 y = 6

y^{^{''}} - 2 y^{^{'}} + 17 y = e x cos (4 x)

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

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