What is the solution for y^{''}+8y=16sin(x) ?

The solution for y^{''}+8y=16sin(x) is y=c_{1}cos(2^{1+1/2}x)+c_{2}sin(2^{1+1/2}x)+16/7 sin(x)

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y^{''}+8y=16sin(x)

y^{^{''}} + 8 y = 16 sin (x)

y = c_{1} cos (2^{1 + \frac{1}{2}} x) + c_{2} sin (2^{1 + \frac{1}{2}} x) + \frac{16}{7} sin (x)

Solution steps

y^{''} (x) + 8 y = 16 sin (x)

Solve linear ODE:

y = c_{1} cos (2^{1 + \frac{1}{2}} x) + c_{2} sin (2^{1 + \frac{1}{2}} x) + \frac{16}{7} sin (x)

y = c_{1} cos (2^{1 + \frac{1}{2}} x) + c_{2} sin (2^{1 + \frac{1}{2}} x) + \frac{16}{7} sin (x)

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

y^{^{''}} + 3 y^{^{'}} = 3

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

y^{^{''}} + 4 y^{^{'}} + 8 y = sin (x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 12 cos (2 x) + 3 sin (2 x)

What is the solution for y^{''}+8y=16sin(x) ?
The solution for y^{''}+8y=16sin(x) is y=c_{1}cos(2^{1+1/2}x)+c_{2}sin(2^{1+1/2}x)+16/7 sin(x)