What is the solution for y^'-ytan(x)=sec(x) ?

The solution for y^'-ytan(x)=sec(x) is y= x/(cos(x))+(c_{1})/(cos(x))

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y^'-ytan(x)=sec(x)

y^{^{'}} - y tan (x) = sec (x)

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

Solution steps

y^{'} (x) - y tan (x) = sec (x)

Solve linear ODE:

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

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

(3 x - 3)^{2} (1 - 5 x)^{4}

3 t + 2

y^{^{''}} + 2 y + y = 0

f (x) = (\frac{5}{x ^{2} + 1}), at x = - 1

y^{^{'}} = 2 e^{(6 x - y)}

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