What is the solution for y^{''}+49y=3 ?

The solution for y^{''}+49y=3 is y=c_{1}cos(7t)+c_{2}sin(7t)+3/49

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y^{''}+49y=3

y^{^{''}} + 49 y = 3

y = c_{1} cos (7 t) + c_{2} sin (7 t) + \frac{3}{49}

Solution steps

y^{''} (t) + 49 y = 3

Solve linear ODE:

y = c_{1} cos (7 t) + c_{2} sin (7 t) + \frac{3}{49}

y = c_{1} cos (7 t) + c_{2} sin (7 t) + \frac{3}{49}

y^{^{''}} - 16 y = sin (4 x)

x^{^{''}} + x = - e^{t}

\frac{d ^{2} y}{d x ^{2}} + 9 y = 5 sin (4 x)

y^{^{''}} + y = cos (t), y (0) = 1, y^{^{'}} (0) = - 3

2 y^{^{''}} + 27 y = e^{t} - e^{- t}, y^{t} (0) = 2, y (0) = 0

What is the solution for y^{''}+49y=3 ?
The solution for y^{''}+49y=3 is y=c_{1}cos(7t)+c_{2}sin(7t)+3/49