Solve the following differential equations?

Y' cosx + y = 1 y' + y secx = secx its a linear differential equation because it is in the form of y' + P(x)y = Q(x). We need to find the integrating factor I(x), which is defined by: I(x) = e^(?P(x) dx), where P(x) = secx. So I(x) = e^(?secx dx) e^(ln(secx + tanx) secx + tanx now multiply secx + tanx to both sides (secx + tanx) y' + y (secx + tanx) secx = secx (tanx + secx) (secx + tanx) y' + y (secx tanx + sec²x) = tanx + sec²x reverse product rule on the left side: d/dx y (secx + tanx) = secx tanx + sec²x integrate both sides: y (secx + tanx) = secx + tanx + C y = 1 + C/(secx + tanx) hope it helps!

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Go to www.wolframalpha.com and enter "solve cos(x)*y'+y=1" to get the answer y(x) = c e^(-2 tanh^(-1)(tan(x/2)))+1 Amazingly, this seems to be Ferroci's answer (appearance of graphs).

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