What is the solution for 9y^'-9tan(x)y=(20)/(cos(x)) ?

The solution for 9y^'-9tan(x)y=(20)/(cos(x)) is y=(20x)/(9cos(x))+(c_{1})/(cos(x))

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9y^'-9tan(x)y=(20)/(cos(x))

9 y^{^{'}} - 9 tan (x) y = \frac{20}{cos ( x )}

y = \frac{20 x}{9 cos ( x )} + \frac{c _{1}}{cos ( x )}

Solution steps

9 y^{'} (x) - 9 tan (x) y = \frac{20}{cos ( x )}

Solve linear ODE:

y = \frac{20 x}{9 cos ( x )} + \frac{c _{1}}{cos ( x )}

y = \frac{20 x}{9 cos ( x )} + \frac{c _{1}}{cos ( x )}

\frac{\partial}{\partial x} (2 x - e^{x + y})

n = 0 \sum \infty \frac{100}{n}

\int x^{- 2} (3 x^{4} + 4 x^{2} - 5) d x

x^{2} + y^{2} = 23

\frac{d x}{d t} = 8 + 4 sin (\frac{t}{4}) - \frac{x}{100}

What is the solution for 9y^'-9tan(x)y=(20)/(cos(x)) ?
The solution for 9y^'-9tan(x)y=(20)/(cos(x)) is y=(20x)/(9cos(x))+(c_{1})/(cos(x))