Solve differential equation dy = (sec x + ytan x)dx.?

Question

&
xdy - ydx + logxdx = 0 at ( x = 1, y = 1)

Answer 1

Both equations can be solved with an integrating factor µ(x)
The appropriate integrating factor for a DE of the form.
dy/dx + p(x) · y = q(x)
is
µ(x) = e^( ∫ p(x) dx)
and leads to the general solution:
y = ( ∫ µ(x)·q(x) dx + c) / µ(x)

(I)
dy = [sec(x) + y·tan(x) ]dx
<=>
dy/dx - tan(x)·y = sec(x)
=>
µ = e^( ∫ -tan(x) dx) = e^( ln(cos(x))) = cos(x)
=>
y = ( ∫ cos(x)·sec(x) dx + c) / cos(x)
= ( ∫ 1 dx + c) / cos(x)
= (x + c) / cos(x)
= (x+c) · sec(x)
c is the constant of integration, which can be found from the boundary condition:
y(x=1) = 1
<=>
(1 + c) · sec(1) = 1
=>
c = cos(1) - 1

=>
y = (x + cos(1) - 1) · sec(x)



(II)
xdy - ydx + ln(x) dx = 0
<=>
dy/dx - (1/x)·y = -ln(x)/x
=>
µ = e^( ∫ -1/x dx) = e^( -ln(x)) = 1/x
=>
y = ( ∫ (1/x) · (-ln(x)/x) dx + c) / (1/x)
= ( ∫ -ln(x)/x² dx + c) · x
substitute t=ln(x) <=> x=e^t => dx = e^t dt
= ( ∫ (-t / (e^t)²) · e^t dt + c) · e^t
= ( ∫ (-t ·e^(-t) dt + c) · e^t
= (+(t+1)·e^(-t) + c) · e^t
= c·e^t + t + 1
= c·x + ln(x) + 1
Apply boundary condition:
y(x=1) = 1
<=>
c·1 + ln(1) + 1 = 1
=>
c = 0
=>

y = ln(x) +1

Answer 2

dy/dx = sec x + y tan x; multiply through by cos x
<=> cos x dy/dx - y sin x = 1
<=> d/dx (y cos x) = 1
<=> y cos x = x + c for some c
<=> y = (x + c) sec x.

x dy/dx - y = -ln x
<=> (1/x) dy/dx - (1/x^2) y = -(ln x)/x^2
<=> d/dx (y/x) = -(ln x)/x^2
<=> y/x = ∫(-ln x) / x^2 dx
Now if we let u = (ln x) / x, we find
du/dx = ((1/x) x - (ln x) (1)) / x^2 = (1 - ln x) / x^2
Hence we have
y/x = ∫-1/x^2 dx + ∫(1-ln x)/x^2 dx
= 1/x + (ln x) / x + c
and so we get
y = 1 + ln x + cx.