What is the solution for y-yy^'+2y=cos(8) ?

The solution for y-yy^'+2y=cos(8) is y=(cos(8)\W(-1/(cos(8)e^{(-9t-c_{1))/(cos(8))}}))/3+(cos(8))/3

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y-yy^'+2y=cos(8)

y - y y^{^{'}} + 2 y = cos (8)

y = \frac{cos ( 8 ) W ( - \frac{1}{c o s ( 8 ) e ^{\frac{- 9 t - c _{1}}{c o s ( 8 )}}} )}{3} + \frac{cos ( 8 )}{3}

Solution steps

y - y y^{'} (t) + 2 y = cos (8)

Solve separable ODE:

y = \frac{cos ( 8 ) W ( - \frac{1}{c o s ( 8 ) e ^{\frac{- 9 t - c _{1}}{c o s ( 8 )}}} )}{3} + \frac{cos ( 8 )}{3}

y = \frac{cos ( 8 ) W ( - \frac{1}{c o s ( 8 ) e ^{\frac{- 9 t - c _{1}}{c o s ( 8 )}}} )}{3} + \frac{cos ( 8 )}{3}

(t^{2} - 24 t + 95) y^{^{'}} = y

\frac{d}{d x} (x^{4} + y^{4}) = \frac{d}{d x} (16)

(x^{2} - 1) \frac{d y}{d x} = y^{2} - 1, y (2) = 2

\frac{d y}{d t} = k y (t) + n

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

What is the solution for y-yy^'+2y=cos(8) ?
The solution for y-yy^'+2y=cos(8) is y=(cos(8)\W(-1/(cos(8)e^{(-9t-c_{1))/(cos(8))}}))/3+(cos(8))/3