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Step-by-Step Solution
Step 1: Rewrite the given differential equation
The given equation is
e^y \frac{dy}{dx} - 2 e^y \sin x + \sin x \cos^2 x = 0
with the initial condition y\left(\frac{\pi}{2}\right) = 0 .
Step 2: Introduce a substitution to simplify the equation
Let t = e^y . Then
\frac{dt}{dx} = e^y \frac{dy}{dx} .
Substituting into the given differential equation:
\frac{dt}{dx} - 2\sin x \, t + \sin x \cos^2 x = 0
can be rearranged to:
\frac{dt}{dx} - 2 \sin x \, t = - \sin x \cos^2 x.
Step 3: Determine the integrating factor (IF)
The differential equation in t is linear in standard form:
\frac{dt}{dx} + P(x)\, t = Q(x),
but here P(x) = -2 \sin x . Hence the integrating factor is given by
\text{IF} = e^{\int P(x)\, dx} = e^{\int -2 \sin x \, dx} = e^{2 \cos x}.
Step 4: Multiply the differential equation by the integrating factor
Multiplying both sides by e^{2\cos x} gives:
e^{2\cos x} \frac{dt}{dx} - 2 \sin x \, t \, e^{2 \cos x} = - \sin x \cos^2 x \, e^{2 \cos x}.
Observe that the left-hand side can be written as the derivative of \bigl(t \, e^{2\cos x}\bigr) :
\frac{d}{dx}\bigl(t \, e^{2\cos x}\bigr) = - \sin x \cos^2 x \, e^{2 \cos x}.
Step 5: Integrate both sides
Integrating, we get
t \, e^{2\cos x} = \int \left( - \sin x \cos^2 x \, e^{2 \cos x} \right) dx + C,
where C is the constant of integration.
Step 6: Express back in terms of y
Recall that t = e^y . Thus
e^y \, e^{2\cos x} = \int \left( - \sin x \cos^2 x \, e^{2 \cos x} \right) dx + C.
Step 7: Use the initial condition to find the constant of integration
We know that y\bigl(\frac{\pi}{2}\bigr) = 0 , which implies e^y = 1 at x = \frac{\pi}{2} and \cos\bigl(\frac{\pi}{2}\bigr) = 0 . Substituting these into the integrated result allows solving for C :
(Details of the integral evaluation and simplification lead to a final expression for the constant, which is found to be \frac{3}{4} .)
Step 8: Solve for e^y and then for y
After evaluating the integral and substituting the constant C = \tfrac{3}{4} , one obtains:
e^y = \frac{1}{2}\cos^2 x - \frac{1}{2} \cos x + \frac{1}{4} + \frac{3}{4} e^{-2\cos x}.
Taking the natural logarithm,
y = \ln \left[ \frac{1}{2}\cos^2 x - \frac{1}{2} \cos x + \frac{1}{4} + \frac{3}{4} e^{-2\cos x} \right].
Step 9: Find y(0)
Substitute x = 0 into the expression for e^y . Recall \cos(0) = 1 :
e^y = \frac{1}{2} (1)^2 - \frac{1}{2} (1) + \frac{1}{4} + \frac{3}{4} e^{-2 \cdot 1} \\
\quad\;\; = \frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{3}{4} e^{-2}
= \frac{1}{4} + \frac{3}{4} e^{-2}.
Step 10: Identify \alpha and \beta and evaluate 4(\alpha + \beta)
We see that
y(0) = \ln \Bigl[ \tfrac{1}{4} + \tfrac{3}{4} e^{-2} \Bigr],
which matches the form y(0) = \ln \bigl(\alpha + \beta e^{-2}\bigr).
Thus we compare directly:
\alpha = \frac{1}{4}, \quad \beta = \frac{3}{4}.
Finally,
4(\alpha + \beta) = 4 \left(\frac{1}{4} + \frac{3}{4}\right) = 4 \times 1 = 4.
Answer
4(\alpha + \beta) = 4.
