What is the solution for y^{''}+9y=2,y^'(0)=5 ?

The solution for y^{''}+9y=2,y^'(0)=5 is y=c_{1}cos(3t)+5/3 sin(3t)+2/9

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y^{''}+9y=2,y^'(0)=5

y^{^{''}} + 9 y = 2, y^{^{'}} (0) = 5

y = c_{1} cos (3 t) + \frac{5}{3} sin (3 t) + \frac{2}{9}

Solution steps

y^{''} (t) + 9 y = 2

Solve linear ODE:

y = c_{1} cos (3 t) + \frac{5}{3} sin (3 t) + \frac{2}{9}

y = c_{1} cos (3 t) + \frac{5}{3} sin (3 t) + \frac{2}{9}

y^{^{''}} + y^{^{'}} - 12 y = sin (t)

y^{^{''}} - y^{^{'}} - 6 y = 250 t e^{- 2 t}

y^{^{''}} - 8 y^{^{'}} + 20 y = 100 x^{2}

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

y^{^{''}} - 8 y^{^{'}} + 16 y = \frac{e ^{4 x}}{x}

What is the solution for y^{''}+9y=2,y^'(0)=5 ?
The solution for y^{''}+9y=2,y^'(0)=5 is y=c_{1}cos(3t)+5/3 sin(3t)+2/9