What is the solution for y^{''}+4y=3x ?

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

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y^{''}+4y=3x

y^{^{''}} + 4 y = 3 x

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

Solution steps

y^{''} (x) + 4 y = 3 x

Solve linear ODE:

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

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

y^{^{''}} + 2 a y^{^{'}} + b^{2} y = sin (c t)

x^{^{''}} + 10 x^{^{'}} + 16 x = 2581 cos (5 t)

(D^{4} + 10 D^{2} + 9) y = sin (3 x) + cos (3 x)

y^{^{''}} + 36 y = 2 e^{6 x}

y^{^{''}} + 2 y^{^{'}} - 8 y = 2 e^{4 t}

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