What is the solution for y^{''}+2y=3x+5 ?

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

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y^{''}+2y=3x+5

y^{^{''}} + 2 y = 3 x + 5

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

Solution steps

y^{''} (x) + 2 y = 3 x + 5

Solve linear ODE:

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

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

y^{^{''}} + a y^{^{'}} + b = 0

y^{^{''}} + k y = t^{2} e^{3 t}, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 2 y^{^{'}} + y = t^{2}, y (0) = 1, y^{^{'}} (0) = 1

y^{^{''}} + 4 y = g (t), y (0) = 3, y^{^{'}} (0) = 0

y^{^{''}} - 12 y^{^{'}} + 36 y = 24 x + 6

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