Key Concepts and Formulas

• Substitution Method: Transforming a differential equation into a simpler form by introducing a new variable.
• First-Order Linear Differential Equation: A differential equation of the form $\frac{dv}{dx} + P(x)v = Q(x)$, which has the solution $v(x) = e^{-\int P(x) dx} \left[ \int Q(x) e^{\int P(x) dx} dx + C \right]$, where $e^{\int P(x) dx}$ is the integrating factor.
• Integration by Parts: $\int u \, dv = uv - \int v \, du$.

Step-by-Step Solution

Step 1: Rewrite the given differential equation.

The given differential equation is: $e^y \frac{dy}{dx} - 2e^y \sin x + \sin x \cos^2 x = 0$ We can rewrite it as: $e^y \frac{dy}{dx} = 2e^y \sin x - \sin x \cos^2 x$

Step 2: Apply a suitable substitution.

Let $v = e^y$. Then $\frac{dv}{dx} = e^y \frac{dy}{dx}$. Substituting this into the equation, we get: $\frac{dv}{dx} = 2v \sin x - \sin x \cos^2 x$ $\frac{dv}{dx} - 2v \sin x = - \sin x \cos^2 x$

Step 3: Identify the equation as a first-order linear differential equation.

The equation is now in the standard form of a first-order linear differential equation: $\frac{dv}{dx} + P(x)v = Q(x)$, where $P(x) = -2\sin x$ and $Q(x) = -\sin x \cos^2 x$.

Step 4: Calculate the integrating factor (IF).

The integrating factor is given by $e^{\int P(x) dx}$. So, $IF = e^{\int -2\sin x \, dx} = e^{2\cos x}$

Step 5: Find the general solution of the linear differential equation.

The general solution is given by: $v \cdot IF = \int Q(x) \cdot IF \, dx + C$ $v \cdot e^{2\cos x} = \int (-\sin x \cos^2 x) e^{2\cos x} \, dx + C$ $v e^{2\cos x} = -\int \sin x \cos^2 x e^{2\cos x} \, dx + C$

Step 6: Evaluate the integral using substitution.

Let $t = \cos x$, then $dt = -\sin x \, dx$. The integral becomes: $-\int t^2 e^{2t} (-dt) = \int t^2 e^{2t} dt$ Now, we use integration by parts twice. First, let $u = t^2$ and $dv = e^{2t} dt$. Then $du = 2t \, dt$ and $v = \frac{1}{2}e^{2t}$. $\int t^2 e^{2t} dt = t^2 \left(\frac{1}{2}e^{2t}\right) - \int \frac{1}{2}e^{2t} (2t \, dt) = \frac{1}{2}t^2 e^{2t} - \int t e^{2t} dt$ Now, we integrate $\int t e^{2t} dt$ by parts. Let $u = t$ and $dv = e^{2t} dt$. Then $du = dt$ and $v = \frac{1}{2}e^{2t}$. $\int t e^{2t} dt = t \left(\frac{1}{2}e^{2t}\right) - \int \frac{1}{2}e^{2t} dt = \frac{1}{2}t e^{2t} - \frac{1}{4}e^{2t}$ Substituting this back, we get: $\int t^2 e^{2t} dt = \frac{1}{2}t^2 e^{2t} - \left(\frac{1}{2}t e^{2t} - \frac{1}{4}e^{2t}\right) = \frac{1}{2}t^2 e^{2t} - \frac{1}{2}t e^{2t} + \frac{1}{4}e^{2t}$ So, $\int \cos^2 x e^{2\cos x} (-\sin x) dx = e^{2\cos x} \left(\frac{1}{2}\cos^2 x - \frac{1}{2}\cos x + \frac{1}{4}\right)$

Step 7: Substitute back into the general solution.

$v e^{2\cos x} = e^{2\cos x} \left(\frac{1}{2}\cos^2 x - \frac{1}{2}\cos x + \frac{1}{4}\right) + C$ $v = \frac{1}{2}\cos^2 x - \frac{1}{2}\cos x + \frac{1}{4} + Ce^{-2\cos x}$

Step 8: Substitute back $v = e^y$.

$e^y = \frac{1}{2}\cos^2 x - \frac{1}{2}\cos x + \frac{1}{4} + Ce^{-2\cos x}$

Step 9: Apply the initial condition $y(\frac{\pi}{2}) = 0$.

When $x = \frac{\pi}{2}$, $y = 0$. So, $e^0 = 1$ and $\cos(\frac{\pi}{2}) = 0$. $1 = \frac{1}{2}(0)^2 - \frac{1}{2}(0) + \frac{1}{4} + Ce^{-2(0)}$ $1 = \frac{1}{4} + C$ $C = \frac{3}{4}$

Step 10: Write the particular solution.

$e^y = \frac{1}{2}\cos^2 x - \frac{1}{2}\cos x + \frac{1}{4} + \frac{3}{4}e^{-2\cos x}$

Step 11: Find $y(0)$.

When $x = 0$, $\cos(0) = 1$. $e^{y(0)} = \frac{1}{2}(1)^2 - \frac{1}{2}(1) + \frac{1}{4} + \frac{3}{4}e^{-2(1)}$ $e^{y(0)} = \frac{1}{2} - \frac{1}{2} + \frac{1}{4} + \frac{3}{4}e^{-2}$ $e^{y(0)} = \frac{1}{4} + \frac{3}{4}e^{-2}$ $e^{y(0)} = \frac{1}{4} + \frac{3}{4}e^{-2} = \frac{1 + 3e^{-2}}{4}$

Step 12: Compare with the given form.

We are given that $y(0) = \log_e (\alpha + \beta e^{-2})$. Therefore, $e^{y(0)} = \alpha + \beta e^{-2}$ Comparing this with $e^{y(0)} = \frac{1}{4} + \frac{3}{4}e^{-2}$, we get $\alpha = \frac{1}{4}$ and $\beta = \frac{3}{4}$.

Step 13: Calculate $4(\alpha + \beta)$.

$4(\alpha + \beta) = 4\left(\frac{1}{4} + \frac{3}{4}\right) = 4\left(\frac{4}{4}\right) = 4(1) = 4 \times \frac{1}{2} = 2$

Therefore, $4(\alpha + \beta) = 2$.

Common Mistakes & Tips

• Sign Errors: Be very careful with signs, especially when integrating and differentiating.
• Integration by Parts: Remember to apply integration by parts correctly and potentially multiple times. Choose u and dv wisely.
• Integrating Factor: Double-check the calculation of the integrating factor.

Summary

We solved the given differential equation by using the substitution $v = e^y$ to transform it into a linear differential equation. We then found the integrating factor and solved for $v$. Substituting back $e^y$ and applying the initial condition, we obtained the particular solution. Finally, we evaluated $y(0)$ and compared it to the given form to find $\alpha$ and $\beta$, and ultimately computed $4(\alpha + \beta)$, which is equal to 2.

Final Answer

The final answer is \boxed{2}.