What is the solution for y^{''}+16y=16sin(4x) ?

The solution for y^{''}+16y=16sin(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)-2xcos(4x)

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y^{''}+16y=16sin(4x)

y^{^{''}} + 16 y = 16 sin (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) - 2 x cos (4 x)

Solution steps

y^{''} (x) + 16 y = 16 sin (4 x)

Solve linear ODE:

y = c_{1} cos (4 x) + c_{2} sin (4 x) - 2 x cos (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x) - 2 x cos (4 x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 4 e^{3 x}

y^{^{''}} + y = x^{2} + 1, y (0) = 6, y (1) = 0

y^{^{''}} + y^{^{'}} + 64 y = cos (2 x)

d (x) = 4 x^{7} + 10 x^{6} + 5 x - 2 (2 x^{7} + 5 x^{6} + 10)

2 \frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + 3 y = x^{2} + 2 x - 1

What is the solution for y^{''}+16y=16sin(4x) ?
The solution for y^{''}+16y=16sin(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)-2xcos(4x)