What is the solution for y^{''}+16y=g(t),y(0)=3,y^'(0)=0 ?

The solution for y^{''}+16y=g(t),y(0)=3,y^'(0)=0 is y=3cos(4t)-(gsin(4t))/(64)+(gt)/(16)

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y^{''}+16y=g(t),y(0)=3,y^'(0)=0

y^{^{''}} + 16 y = g (t), y (0) = 3, y^{^{'}} (0) = 0

y = 3 cos (4 t) - \frac{g sin ( 4 t )}{64} + \frac{g t}{16}

Solution steps

y^{''} (t) + 16 y = g (t)

Solve linear ODE:

y = 3 cos (4 t) - \frac{g sin ( 4 t )}{64} + \frac{g t}{16}

y = 3 cos (4 t) - \frac{g sin ( 4 t )}{64} + \frac{g t}{16}

\frac{d ^{2} y}{d x ^{2}} - 3 \frac{d y}{d x} + 2 y = x^{2} + 8 x + 9

y^{^{''}} + 4 y^{^{'}} + 4 y = 1

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

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

y^{^{''}} - 5 y^{^{'}} - 6 y = - sin (2 x)

What is the solution for y^{''}+16y=g(t),y(0)=3,y^'(0)=0 ?
The solution for y^{''}+16y=g(t),y(0)=3,y^'(0)=0 is y=3cos(4t)-(gsin(4t))/(64)+(gt)/(16)