Need help with this 1st order linear diff. equation!!?

Let u = 1/cos(x) = sec(x) so u' = tan(x)sec(x)

You just have to realize that tan(x) = u'/u so the deq becomes

y' + (u'/u)y = u

Now multiply by u

uy' + u'y = u^2 = sec^2(x)

The LHS is just the product rule applied to uy

d(uy) = sec^2(x)dx

Integrating we get

uy = tan(x) + C
sec(x)y = tan(x) + C
y = tan(x)/sec(x) + C/sec(x)

y = tan(x)cos(x) + Ccos(x)

y = sin(x) + Ccos(x)

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