What is the solution for y^'+36ycot(6x)=42cos(6x) ?

The solution for y^'+36ycot(6x)=42cos(6x) is y=sin(6x)+(c_{1})/(sin^6(6x))

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y^'+36ycot(6x)=42cos(6x)

y^{^{'}} + 36 y cot (6 x) = 42 cos (6 x)

y = sin (6 x) + \frac{c _{1}}{sin ^{6} ( 6 x )}

Solution steps

y^{'} (x) + 36 y cot (6 x) = 42 cos (6 x)

Solve linear ODE:

y = sin (6 x) + \frac{c _{1}}{sin ^{6} ( 6 x )}

y = sin (6 x) + \frac{c _{1}}{sin ^{6} ( 6 x )}

\frac{( x ^{2} )}{1 + x}

\int 6 x (4 + x^{2})^{\frac{1}{2}} d x

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

- 3 sin (4 t)

x \cdot \frac{d y}{d x} - x^{2} + x \cdot e^{- x} = 0

What is the solution for y^'+36ycot(6x)=42cos(6x) ?
The solution for y^'+36ycot(6x)=42cos(6x) is y=sin(6x)+(c_{1})/(sin^6(6x))