Key Concepts and Formulas

• Separable Differential Equations: A differential equation of the form $M(x)dx + N(y)dy = 0$ can be solved by integrating both sides separately.
• Integration Techniques: Knowledge of basic integration rules, including substitution.
• Trigonometric Identities: The tangent subtraction formula: $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.

Step-by-Step Solution

Step 1: Separating the Variables

We are given the differential equation $(1+y^2)(1+\log_e x)dx + xdy = 0$ Our goal is to rewrite this equation in the form $M(x)dx + N(y)dy = 0$. First, subtract $xdy$ from both sides: $(1+y^2)(1+\log_e x)dx = -xdy$ Now, divide both sides by $x(1+y^2)$ to separate the variables: $\frac{1+\log_e x}{x}dx = -\frac{1}{1+y^2}dy$ Rearranging the terms, we get $\frac{1+\log_e x}{x}dx + \frac{1}{1+y^2}dy = 0$ This is now in the desired separable form.

Step 2: Integrating Both Sides

Integrate both sides of the separated equation: $\int \frac{1+\log_e x}{x}dx + \int \frac{1}{1+y^2}dy = C$ where $C$ is the constant of integration. Let's evaluate the integrals separately. For the first integral, we have $\int \frac{1+\log_e x}{x}dx = \int \frac{1}{x}dx + \int \frac{\log_e x}{x}dx$ The first term integrates to $\log_e x$. For the second term, let $u = \log_e x$, then $du = \frac{1}{x}dx$. So, $\int \frac{\log_e x}{x}dx = \int u du = \frac{u^2}{2} = \frac{(\log_e x)^2}{2}$ Thus, $\int \frac{1+\log_e x}{x}dx = \log_e x + \frac{(\log_e x)^2}{2}$ The second integral is a standard integral: $\int \frac{1}{1+y^2}dy = \arctan(y)$ Combining these results, we have the general solution: $\log_e x + \frac{(\log_e x)^2}{2} + \arctan(y) = C$

Step 3: Applying the Initial Condition

We are given that the solution passes through the point $(1,1)$. This means when $x=1$, $y=1$. Substituting these values into the general solution: $\log_e 1 + \frac{(\log_e 1)^2}{2} + \arctan(1) = C$ Since $\log_e 1 = 0$ and $\arctan(1) = \frac{\pi}{4}$, we have $0 + \frac{0^2}{2} + \frac{\pi}{4} = C$ $C = \frac{\pi}{4}$ So the particular solution is $\log_e x + \frac{(\log_e x)^2}{2} + \arctan(y) = \frac{\pi}{4}$

Step 4: Evaluating y(e)

We want to find $y(e)$, which means we want to find $y$ when $x=e$. Substitute $x=e$ into the particular solution: $\log_e e + \frac{(\log_e e)^2}{2} + \arctan(y(e)) = \frac{\pi}{4}$ Since $\log_e e = 1$, we have $1 + \frac{1^2}{2} + \arctan(y(e)) = \frac{\pi}{4}$ $\frac{3}{2} + \arctan(y(e)) = \frac{\pi}{4}$ $\arctan(y(e)) = \frac{\pi}{4} - \frac{3}{2}$ Taking the tangent of both sides: $y(e) = \tan\left(\frac{\pi}{4} - \frac{3}{2}\right)$

Step 5: Simplifying y(e) and Comparing

Using the tangent subtraction formula, $\tan(A-B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$, with $A = \frac{\pi}{4}$ and $B = \frac{3}{2}$, we get $y(e) = \frac{\tan(\frac{\pi}{4}) - \tan(\frac{3}{2})}{1 + \tan(\frac{\pi}{4})\tan(\frac{3}{2})}$ Since $\tan(\frac{\pi}{4}) = 1$, we have $y(e) = \frac{1 - \tan(\frac{3}{2})}{1 + \tan(\frac{3}{2})}$ Comparing this with the given form $y(e) = \frac{\alpha - \tan(\frac{3}{2})}{\beta + \tan(\frac{3}{2})}$, we can identify $\alpha = 1$ and $\beta = 1$.

Step 6: Calculating α + 2β

Finally, we calculate $\alpha + 2\beta$: $\alpha + 2\beta = 1 + 2(1) = 1 + 2 = 3$

Common Mistakes & Tips

• Sign Errors: Be careful with signs when separating variables and applying the tangent subtraction formula.
• Integration Constant: Always remember the constant of integration and solve for it using the initial condition.
• Trigonometric Identities: Make sure to correctly apply the tangent subtraction formula.

Summary

We solved the given differential equation by separating variables, integrating both sides, and applying the initial condition to find the particular solution. We then evaluated the solution at $x=e$ and used the tangent subtraction formula to express the result in the desired form. Finally, we calculated $\alpha + 2\beta$ and found it to be 3.

The final answer is \boxed{3}.