What is the solution for y^{''}+4y=2x ?

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

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

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

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x}{2}

y = c_{1} cos (2 x) + c_{2} sin (2 x) + \frac{x}{2}

y^{^{'''}} - 6 y^{^{''}} + 13 y^{^{'}} - 10 y = e^{2 x}

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

y^{^{''}} - 10 y^{^{'}} + 25 y = x + 1 + e^{5 x}

y^{^{''}} - y^{^{'}} = - 7

x^{^{''}} + 9 x = sin (2.9 t)

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