What is the solution for y^{''}+16y=9(t)-9(t-98pi),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+16y=9(t)-9(t-98pi),y(0)=0,y^'(0)=0 is y=-(441pi)/8 cos(4t)+(441pi)/8

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y^{''}+16y=9(t)-9(t-98pi),y(0)=0,y^'(0)=0

y^{^{''}} + 16 y = 9 (t) - 9 (t - 98 π), y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{441 π}{8} cos (4 t) + \frac{441 π}{8}

Solution steps

y^{''} (t) + 16 y = 9 (t) - 9 (t - 98 π)

Solve linear ODE:

y = - \frac{441 π}{8} cos (4 t) + \frac{441 π}{8}

y = - \frac{441 π}{8} cos (4 t) + \frac{441 π}{8}

y^{^{''}} - 4 y^{^{'}} + 3 y = 8 e^{- x} + 9

y^{^{''}} + 18 y^{^{'}} + 82 y = 2 e^{- 9 x} cos (x)

y^{^{''}} + 9 y^{^{'}} = t^{2} e^{3 t} + 6

\frac{d ^{2} y}{d t ^{2}} - 12 \frac{d y}{d t} + 36 y = 7 + 4 t

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

What is the solution for y^{''}+16y=9(t)-9(t-98pi),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+16y=9(t)-9(t-98pi),y(0)=0,y^'(0)=0 is y=-(441pi)/8 cos(4t)+(441pi)/8