What is the solution for y^{''}+16y=32t^2-48t+100,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+16y=32t^2-48t+100,y(0)=0,y^'(0)=1 is y=-6cos(4t)+sin(4t)+2t^2-3t+6

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y^{''}+16y=32t^2-48t+100,y(0)=0,y^'(0)=1

y^{^{''}} + 16 y = 32 t^{2} - 48 t + 100, y (0) = 0, y^{^{'}} (0) = 1

y = - 6 cos (4 t) + sin (4 t) + 2 t^{2} - 3 t + 6

Solution steps

y^{''} (t) + 16 y = 32 t^{2} - 48 t + 100

Solve linear ODE:

y = - 6 cos (4 t) + sin (4 t) + 2 t^{2} - 3 t + 6

y = - 6 cos (4 t) + sin (4 t) + 2 t^{2} - 3 t + 6

y^{^{''}} + y^{^{'}} = cos (x), y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} - 9 y = cosh (3 x)

y^{^{''''}} - 16 y = - 12 t^{3}

y^{^{''''}} - 16 y = 75 sin (t)

y^{^{''}} - 100 y = sinh (10 x)

What is the solution for y^{''}+16y=32t^2-48t+100,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+16y=32t^2-48t+100,y(0)=0,y^'(0)=1 is y=-6cos(4t)+sin(4t)+2t^2-3t+6