What is the solution for y^{''}+3y=2x ?

The solution for y^{''}+3y=2x is y=c_{1}cos(sqrt(3)x)+c_{2}sin(sqrt(3)x)+(2x)/3

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y^{''}+3y=2x

y^{^{''}} + 3 y = 2 x

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x}{3}

y = c_{1} cos (3 x) + c_{2} sin (3 x) + \frac{2 x}{3}

y^{^{''}} + 16 y = cos (4 t), y (0) = 0, y^{^{'}} (0) = 1

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

y^{^{''}} + 2 y^{^{'}} - 15 y = - 53 t + 30 t^{2}

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{x}, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = x^{(- 3)} e^{3} x

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