What is the solution for y^{''}+36y=sin(4x) ?

The solution for y^{''}+36y=sin(4x) is y=c_{1}cos(6x)+c_{2}sin(6x)+1/20 sin(4x)

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y^{''}+36y=sin(4x)

y^{^{''}} + 36 y = sin (4 x)

y = c_{1} cos (6 x) + c_{2} sin (6 x) + \frac{1}{20} sin (4 x)

Solution steps

y^{''} (x) + 36 y = sin (4 x)

Solve linear ODE:

y = c_{1} cos (6 x) + c_{2} sin (6 x) + \frac{1}{20} sin (4 x)

y = c_{1} cos (6 x) + c_{2} sin (6 x) + \frac{1}{20} sin (4 x)

y^{^{''}} + 3 y = - 48 x^{2} \cdot e^{3 x}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 14 y = 7 + 4 x

y^{^{''}} - 3 y + 2 y = sin (e^{- x})

y^{^{'''}} + 16 y^{^{'}} = t

y^{^{''}} + 3 y^{^{'}} + 2 y = \frac{x}{e ^{x}} + \frac{x ^{2}}{e ^{2 x}}

What is the solution for y^{''}+36y=sin(4x) ?
The solution for y^{''}+36y=sin(4x) is y=c_{1}cos(6x)+c_{2}sin(6x)+1/20 sin(4x)