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Step-by-Step Solution
Step 1: Identify the differential equation and the initial condition
We are given the differential equation:
$
\frac{dy}{dx} = \frac{\tan x + y}{\sin x \bigl(\sec x - \sin x \tan x\bigr)},
$
valid for
$
x \in\!\bigl(0, \tfrac{\pi}{2}\bigr),
$
along with the condition
$
y\bigl(\tfrac{\pi}{4}\bigr) = 2.
$
Step 2: Simplify the denominator
Note that
$
\sec x = \frac{1}{\cos x}
$
and
$
\tan x = \frac{\sin x}{\cos x}.
$
Therefore,
$
\sec x - \sin x \tan x
= \frac{1}{\cos x} - \sin x \cdot \frac{\sin x}{\cos x}
= \frac{1 - \sin^2 x}{\cos x}
= \frac{\cos^2 x}{\cos x}
= \cos x.
$
Substituting this back into the differential equation gives:
$
\frac{dy}{dx}
= \frac{\tan x + y}{\sin x \cos x}.
$
Step 3: Rewrite in a more convenient form
Since
$
\tan x = \frac{\sin x}{\cos x},
$
we write
$
\frac{dy}{dx}
= \frac{\tfrac{\sin x}{\cos x} + y}{\sin x \cos x}
= \frac{\sin x + y \cos x}{\sin x \cos^2 x}.
$
Our goal is to bring this into the standard linear differential equation form.
Step 4: Express as a linear differential equation
Separate the terms:
$
\frac{dy}{dx}
= \frac{\sin x}{\sin x \cos^2 x} + y \cdot \frac{\cos x}{\sin x \cos^2 x}
= \frac{1}{\cos^2 x} + \frac{y}{\sin x \cos x}.
$
Rearranging:
$
\frac{dy}{dx} - \frac{1}{\sin x \cos x}\,y
= \frac{1}{\cos^2 x}.
$
This is a linear differential equation of the form
$
\frac{dy}{dx} + p(x)\,y = q(x),
$
where
$
p(x) = -\frac{1}{\sin x \cos x}
$
and
$
q(x) = \frac{1}{\cos^2 x}.
$
Step 5: Find the integrating factor
The integrating factor (I.F.) is
$
\exp \Bigl(\int p(x)\,dx \Bigr)
= \exp \Bigl(-\int \frac{1}{\sin x \cos x}\,dx \Bigr).
$
Since
$
\frac{1}{\sin x \cos x} = \frac{2}{\sin(2x)},
$
we get
$
\int p(x)\,dx
= -\int \frac{2}{\sin(2x)}\,dx.
$
Let
$
u = 2x \implies du = 2\,dx \implies dx = \frac{du}{2}.
$
Thus:
$
-\int \frac{2}{\sin(2x)}\,dx
= -\int \frac{2}{\sin(u)} \cdot \frac{du}{2}
= -\int \csc(u)\,du.
$
We know
$
\int \csc(u)\,du = \ln|\csc(u) - \cot(u)|.
$
So:
$
-\int \csc(u)\,du = -\ln|\csc(u) - \cot(u)|.
$
Over
$
x \in\bigl(0,\tfrac{\pi}{2}\bigr),
$
this effectively simplifies to an integrating factor
$
\frac{1}{\tan x}.
$
Step 6: Multiply the equation by the integrating factor
By multiplying both sides of
$
\frac{dy}{dx} - \frac{1}{\sin x \cos x}\,y = \frac{1}{\cos^2 x}
$
by
$
\frac{1}{\tan x},
$
we form the derivative of
$
\frac{y}{\tan x}.
$
In general, for a linear differential equation, we have:
$
\bigl(\text{I.F.} \times y\bigr)' = \text{I.F.} \cdot q(x).
$
Hence:
$
\Bigl(\frac{y}{\tan x}\Bigr)' = \frac{1}{\tan x} \cdot \frac{1}{\cos^2 x}.
$
Step 7: Integrate both sides
Let
$
t = \tan x.
$
Then
$
dt = \sec^2 x \,dx = \frac{1}{\cos^2 x}\,dx.
$
Therefore,
$
\int \frac{1}{\tan x} \cdot \frac{1}{\cos^2 x}\,dx
= \int \frac{1}{t}\,dt
= \ln|t| + C
= \ln(\tan x) + C,
$
since
$
\tan x > 0
$
for
$
x\in\bigl(0,\tfrac{\pi}{2}\bigr).
$
Hence,
$
\frac{y}{\tan x} = \ln(\tan x) + C.
$
This gives the general solution:
$
y = \tan x\,\Bigl[\ln(\tan x) + C\Bigr].
$
Step 8: Apply the initial condition to find C
We have
$
y\bigl(\tfrac{\pi}{4}\bigr) = 2.
$
Since
$
\tan\bigl(\tfrac{\pi}{4}\bigr) = 1,
$
we get
$
2 = 1 \times \bigl[\ln(1) + C\bigr] = 0 + C,
$
so
$
C = 2.
$
Step 9: Write the particular solution and find $y\bigl(\frac{\pi}{3}\bigr)$
The particular solution becomes
$
y = \tan x \,\Bigl[\ln(\tan x) + 2\Bigr].
$
We need
$
y\bigl(\tfrac{\pi}{3}\bigr).
$
Since
$
\tan\bigl(\tfrac{\pi}{3}\bigr) = \sqrt{3},
$
it follows:
$
y\bigl(\tfrac{\pi}{3}\bigr) = \sqrt{3}\Bigl[\ln\bigl(\sqrt{3}\bigr) + 2\Bigr].
$
Recall
$
\ln\bigl(\sqrt{3}\bigr) = \frac{1}{2} \ln(3).
$
Therefore,
$
y\bigl(\tfrac{\pi}{3}\bigr)
= \sqrt{3}\Bigl(2 + \ln(\sqrt{3})\Bigr).
$
Final Answer
$
\sqrt{3}\Bigl(2 + \ln\bigl(\sqrt{3}\bigr)\Bigr).
$
