What is the solution for y^{''}+4y=16sin(2t),y(0)=0,y^'(0)=0 ?

The solution for y^{''}+4y=16sin(2t),y(0)=0,y^'(0)=0 is y=-4tcos(2t)+2sin(2t)

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y^{''}+4y=16sin(2t),y(0)=0,y^'(0)=0

y^{^{''}} + 4 y = 16 sin (2 t), y (0) = 0, y^{^{'}} (0) = 0

y = - 4 t cos (2 t) + 2 sin (2 t)

Solution steps

y^{''} (t) + 4 y = 16 sin (2 t)

Solve linear ODE:

y = - 4 t cos (2 t) + 2 sin (2 t)

y = - 4 t cos (2 t) + 2 sin (2 t)

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

\frac{d ^{2} x}{d t ^{2}} - \frac{d x}{d t} - 2 x = 4 t

x^{^{''}} + 2 x + 5 x = 1, x^{^{'}} (0) = 0, x (0) = 0

(d^{2} - 4 d + 4) y = x^{2} + e^{x} + cos (2 x)

y^{^{''}} + y^{^{'}} - 6 y = - e^{2 x} + 12 x^{2} + 2 x + 3

What is the solution for y^{''}+4y=16sin(2t),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=16sin(2t),y(0)=0,y^'(0)=0 is y=-4tcos(2t)+2sin(2t)