Question

Let \(y=y(x)\) be solution of the following differential equation \(e^y \frac{dy}{dx} - 2e^y \sin x + \sin x \cos^2 x = 0, y(\pi/2) = 0\). If \(y(0) = \log_e(\alpha + \beta e^{-2})\), then \(4(\alpha + \beta)\) is equal to __________.

Hint:

For integrals of the form $\int x^n e^x dx$, use the formula $e^x [x^n - nx^{n-1} + n(n-1)x^{n-2} - .......]$. It's a much faster way to integrate by parts multiple times.

Answer 1

Correct Answer: 4

Step 1: Understanding the Concept:
This is a first-order differential equation that can be transformed into a linear differential equation by substituting \(v = e^y\).
Once the general solution is found, we apply initial conditions to determine constants and evaluate the function at a specific point.
Step 2: Key Formula or Approach:
1. Substitution: \(v = e^y \implies \frac{dv}{dx} = e^y \frac{dy}{dx}\).
2. Linear DE: \(\frac{dv}{dx} + P(x)v = Q(x)\).
3. Integrating Factor: \(IF = e^{\int P(x) dx}\).
Step 3: Detailed Explanation:
The equation is \(e^y \frac{dy}{dx} - 2 e^y \sin x = -\sin x \cos^2 x\).
Let \(v = e^y \implies v' = e^y y'\).
The DE becomes: \(\frac{dv}{dx} - (2 \sin x)v = -\sin x \cos^2 x\).
Calculate \(IF = e^{\int -2 \sin x dx} = e^{2 \cos x}\).
The solution is:
\[ v \cdot e^{2 \cos x} = \int -\sin x \cos^2 x \cdot e^{2 \cos x} dx \]
Let \(2 \cos x = u \implies -2 \sin x dx = du \implies -\sin x dx = \frac{1}{2} du\).
Also \(\cos x = u/2\), so \(\cos^2 x = u^2/4\).
Integral \(= \int \frac{u^2}{4} e^u \cdot \frac{1}{2} du = \frac{1}{8} \int u^2 e^u du = \frac{1}{8} e^u(u^2 - 2u + 2) + C\).
Substituting back:
\[ e^y e^{2 \cos x} = \frac{1}{8} e^{2 \cos x} (4 \cos^2 x - 4 \cos x + 2) + C \]
\[ e^y = \frac{1}{2} \cos^2 x - \frac{1}{2} \cos x + \frac{1}{4} + C e^{-2 \cos x} \]
Apply \(y(\pi/2) = 0\):
\(1 = 0 - 0 + 1/4 + C e^0 \implies C = 3/4\).
So, \(e^y = \frac{1}{2} \cos^2 x - \frac{1}{2} \cos x + \frac{1}{4} + \frac{3}{4} e^{-2 \cos x}\).
At \(x=0\):
\(e^{y(0)} = \frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{3}{4} e^{-2} = \frac{1}{4} + \frac{3}{4} e^{-2}\).
Given \(y(0) = \ln(\alpha + \beta e^{-2})\), we get \(\alpha = 1/4, \beta = 3/4\).
Sum \(\alpha + \beta = 1\), so \(4(\alpha + \beta) = 4\).
Step 4: Final Answer:
The result is 4.