What is the solution for y^'+10(tan(10x))y=9cos(10x) ?

The solution for y^'+10(tan(10x))y=9cos(10x) is y=(9x)/(sec(10x))+(c_{1})/(sec(10x))

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y^'+10(tan(10x))y=9cos(10x)

y^{^{'}} + 10 (tan (10 x)) y = 9 cos (10 x)

y = \frac{9 x}{sec ( 10 x )} + \frac{c _{1}}{sec ( 10 x )}

Solution steps

y^{'} (x) + 10 (tan (10 x)) y = 9 cos (10 x)

Solve linear ODE:

y = \frac{9 x}{sec ( 10 x )} + \frac{c _{1}}{sec ( 10 x )}

y = \frac{9 x}{sec ( 10 x )} + \frac{c _{1}}{sec ( 10 x )}

\int (3 x^{2} + 4) x - 1 d x

\frac{d x}{d t} = - \frac{1}{2} x + t

\frac{d}{d t} (7 sin^{3} (t))

\int_{4}^{7} (x - 5)^{2} d x

y^{^{'}} + y = e^{x}, y (0) = 6

What is the solution for y^'+10(tan(10x))y=9cos(10x) ?
The solution for y^'+10(tan(10x))y=9cos(10x) is y=(9x)/(sec(10x))+(c_{1})/(sec(10x))