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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''''}} - 8 y^{^{''}} + 16 y = 9 e^{2 x}

y^{^{''}} - 4 y = (x^{2} - 6) sin (2 x)

y^{^{''}} - y = - 6 sin (x), y (0) = 3, y^{^{'}} (0) = 2

y^{^{''}} - 2 y^{^{'}} + 5 y = e^{x} \cdot cos (2 x)

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

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