What is the solution for y^{''}+3y^'=(10sin(7t))/2 ?

The solution for y^{''}+3y^'=(10sin(7t))/2 is y=c_{1}+c_{2}e^{-3t}-5/58 sin(7t)-15/406 cos(7t)

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y^{''}+3y^'=(10sin(7t))/2

y^{^{''}} + 3 y^{^{'}} = \frac{10 sin ( 7 t )}{2}

y = c_{1} + c_{2} e^{- 3 t} - \frac{5}{58} sin (7 t) - \frac{15}{406} cos (7 t)

Solution steps

y^{''} (t) + 3 y^{'} (t) = \frac{10 sin ( 7 t )}{2}

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 t} - \frac{5}{58} sin (7 t) - \frac{15}{406} cos (7 t)

y = c_{1} + c_{2} e^{- 3 t} - \frac{5}{58} sin (7 t) - \frac{15}{406} cos (7 t)

y^{^{''}} - 4 y^{^{'}} + 4 y = 2 x

y^{^{''}} - 4 y^{^{'}} + 3 y = \frac{e ^{t}}{1 + e ^{t}}

\frac{d ^{2} y}{d t ^{2}} + 9 y = 12 cos (3 t)

y^{^{''}} + y = 5 t^{2} - 9, y (0) = 0, y^{^{'}} (0) = - 2

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

What is the solution for y^{''}+3y^'=(10sin(7t))/2 ?
The solution for y^{''}+3y^'=(10sin(7t))/2 is y=c_{1}+c_{2}e^{-3t}-5/58 sin(7t)-15/406 cos(7t)