What is the solution for 2y^{''}+7y=-8/3 x ?

The solution for 2y^{''}+7y=-8/3 x is y=c_{1}cos(sqrt(7/2)x)+c_{2}sin(sqrt(7/2)x)-(8x)/(21)

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2y^{''}+7y=-8/3 x

2 y^{^{''}} + 7 y = - \frac{8}{3} x

y = c_{1} cos (\frac{7}{2} x) + c_{2} sin (\frac{7}{2} x) - \frac{8 x}{21}

Solution steps

2 y^{''} (x) + 7 y = - \frac{8}{3} x

Solve linear ODE:

y = c_{1} cos (\frac{7}{2} x) + c_{2} sin (\frac{7}{2} x) - \frac{8 x}{21}

y = c_{1} cos (\frac{7}{2} x) + c_{2} sin (\frac{7}{2} x) - \frac{8 x}{21}

y^{^{''}} + y = 5 e^{- 3 x}, y (0) = 0, y^{^{'}} (0) = 0

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

y^{^{''}} + 4 y = sin (2 x) + 1

x^{^{''}} + 8 x^{^{'}} + 16 x = 6 sin (4 t)

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

What is the solution for 2y^{''}+7y=-8/3 x ?
The solution for 2y^{''}+7y=-8/3 x is y=c_{1}cos(sqrt(7/2)x)+c_{2}sin(sqrt(7/2)x)-(8x)/(21)