Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dv}{dy} + P(y)v = Q(y)$ has a general solution given by $v \cdot (\text{Integrating Factor}) = \int Q(y) \cdot (\text{Integrating Factor}) \, dy + C$.
• Integrating Factor (IF): For the linear differential equation above, the integrating factor is $e^{\int P(y) \, dy}$.
• Substitution: Using a substitution to transform a complex differential equation into a simpler, solvable form.

Step-by-Step Solution

Step 1: Rewrite the given differential equation.

The given differential equation is $(1+\log _{e} x) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$ We can rewrite this as: $\frac{dx}{dy} - \frac{x \log_e x}{1 + \log_e x} = \frac{e^y}{1 + \log_e x}$

Step 2: Perform a suitable substitution.

Let $v = \log_e x$. Then, $x = e^v$ and $\frac{dx}{dv} = e^v$. Also, $\frac{dv}{dx} = \frac{1}{x}$. Thus, $\frac{dx}{dy} = \frac{dx}{dv} \cdot \frac{dv}{dy} = e^v \frac{dv}{dy}$. Substituting into the differential equation: $e^v \frac{dv}{dy} - \frac{e^v v}{1+v} = \frac{e^y}{1+v}$ Dividing by $e^v$: $\frac{dv}{dy} - \frac{v}{1+v} = \frac{e^{y-v}}{1+v}$

This form is not linear. Let's go back to the original equation and try dividing by $(1 + \log_e x)$: $\frac{dx}{dy} - \frac{x \log_e x}{1 + \log_e x} = \frac{e^y}{1 + \log_e x}$ Let $z = x \log_e x$. Then $\frac{dz}{dx} = \log_e x + x \cdot \frac{1}{x} = \log_e x + 1$. Therefore, $\frac{dy}{dx} = \frac{1 + \log_e x}{x \log_e x + e^y}$. Then $\frac{dx}{dy} = \frac{x \log_e x + e^y}{1 + \log_e x} = \frac{x \log_e x}{1 + \log_e x} + \frac{e^y}{1 + \log_e x}$. We see that this matches the original equation.

Let's try a different approach. The equation is $(1+\log _{e} x) \frac{d x}{d y}-x \log _{e} x=e^{y}$ $\frac{dx}{dy} - \frac{x \log x}{1 + \log x} = \frac{e^y}{1 + \log x}$ $\frac{dy}{dx} = \frac{1 + \log x}{x \log x + e^y}$ This doesn't seem to simplify easily. Let's reconsider the original equation and rewrite it as: $(1+\log x) dx - (x \log x + e^y) dy = 0$ $(1+\log x) dx = (x \log x + e^y) dy$ $\frac{dx}{dy} = \frac{x \log x + e^y}{1 + \log x}$ $\frac{dx}{dy} - \frac{x \log x}{1 + \log x} = \frac{e^y}{1 + \log x}$ Let $t = \log x$. Then $x = e^t$, and $\frac{dx}{dt} = e^t$, so $\frac{dt}{dx} = \frac{1}{x} = e^{-t}$. Thus $\frac{dx}{dy} = \frac{dx}{dt} \frac{dt}{dy} = e^t \frac{dt}{dy}$. Substituting, we have $e^t \frac{dt}{dy} - \frac{e^t t}{1+t} = \frac{e^y}{1+t}$ $\frac{dt}{dy} - \frac{t}{1+t} = \frac{e^{y-t}}{1+t}$

Step 3: Another Substitution.

Let $u = x\log x$. Then $\frac{du}{dx} = \log x + x\cdot \frac{1}{x} = \log x + 1$. The differential equation becomes $(1+\log x)\frac{dx}{dy} = x\log x + e^y$ $\frac{du}{dy} = x\log x + e^y = u + e^y$ $\frac{du}{dy} - u = e^y$

Step 4: Solve the linear differential equation.

This is a first-order linear differential equation in the form $\frac{du}{dy} + P(y)u = Q(y)$ where $P(y) = -1$ and $Q(y) = e^y$. The integrating factor is $IF = e^{\int P(y) dy} = e^{\int -1 dy} = e^{-y}$. The solution is given by $u \cdot IF = \int Q(y) \cdot IF \, dy + C$ $u e^{-y} = \int e^y e^{-y} dy + C$ $u e^{-y} = \int 1 dy + C$ $u e^{-y} = y + C$ $u = (y+C)e^y$

Step 5: Substitute back to find the solution in terms of x and y.

Since $u = x \log x$, we have $x \log x = (y+C)e^y$.

Step 6: Use the initial condition (1, 0) to find C.

Given that the solution curve passes through $(1, 0)$, we have $x = 1$ and $y = 0$. Substituting these values into the equation $x \log x = (y+C)e^y$, we get $1 \cdot \log 1 = (0+C)e^0$ $0 = C \cdot 1$ $C = 0$ Thus, the particular solution is $x \log x = y e^y$.

Step 7: Use the point ($\alpha$, 2) to find $\alpha$.

Given that the solution curve passes through $(\alpha, 2)$, we have $x = \alpha$ and $y = 2$. Substituting these values into the equation $x \log x = y e^y$, we get $\alpha \log \alpha = 2 e^2$ $\log (\alpha^\alpha) = 2 e^2$ $\alpha^\alpha = e^{2e^2}$

Step 8: Find the value of $\alpha^\alpha$.

We have found that $\alpha^\alpha = e^{2e^2}$.

Common Mistakes & Tips

• Carefully choose the correct substitution to simplify the differential equation.
• Remember to find the integrating factor correctly and apply it to solve the linear differential equation.
• Don't forget to substitute back to the original variables after solving the equation.

Summary

We solved the given differential equation by using the substitution $u = x \log x$ to transform it into a linear differential equation. We found the integrating factor and solved for $u$ in terms of $y$. Then, we substituted back to get the solution in terms of $x$ and $y$. Using the initial condition $(1, 0)$, we found the constant $C = 0$. Finally, using the point $(\alpha, 2)$, we found that $\alpha^\alpha = e^{2e^2}$.

Final Answer

The final answer is \boxed{e^{2 e^{2}}}, which corresponds to option (D).