Key Concepts and Formulas

• Linear First-Order Differential Equation: A differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y)$ and $Q(y)$ are functions of $y$ only.
• Integrating Factor (I.F.): For a linear first-order differential equation in the form $\frac{dx}{dy} + P(y)x = Q(y)$, the integrating factor is given by I.F. $= e^{\int P(y) dy}$.
• General Solution: The general solution of the linear first-order differential equation is given by $x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C$, where $C$ is the constant of integration.

Step-by-Step Solution

Step 1: Rearrange the Differential Equation into Standard Linear Form

We are given the differential equation: $(1+y^2) + (x - 2e^{\tan^{-1}y})\frac{dy}{dx} = 0$ Since we are given that $x=f(y)$, it suggests rewriting the equation in terms of $\frac{dx}{dy}$. We multiply the entire equation by $\frac{dx}{dy}$: $(1+y^2)\frac{dx}{dy} + (x - 2e^{\tan^{-1}y})\frac{dy}{dx}\frac{dx}{dy} = 0$ Since $\frac{dy}{dx}\frac{dx}{dy} = 1$, we have: $(1+y^2)\frac{dx}{dy} + x - 2e^{\tan^{-1}y} = 0$ Now, isolate the $\frac{dx}{dy}$ term and move other terms to the right-hand side: $(1+y^2)\frac{dx}{dy} + x = 2e^{\tan^{-1}y}$ Divide by $(1+y^2)$ to get the equation in the standard form: $\frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{2e^{\tan^{-1}y}}{1+y^2}$ Comparing this with $\frac{dx}{dy} + P(y)x = Q(y)$, we have: $P(y) = \frac{1}{1+y^2} \quad \text{and} \quad Q(y) = \frac{2e^{\tan^{-1}y}}{1+y^2}$

Step 2: Calculate the Integrating Factor (I.F.)

The integrating factor is given by: $\text{I.F.} = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy}$ Since $\int \frac{1}{1+y^2} dy = \tan^{-1}y$, we have: $\text{I.F.} = e^{\tan^{-1}y}$

Step 3: Find the General Solution

The general solution is given by: $x \cdot (\text{I.F.}) = \int Q(y) \cdot (\text{I.F.}) dy + C$ Substituting the values of I.F. and $Q(y)$, we have: $x e^{\tan^{-1}y} = \int \frac{2e^{\tan^{-1}y}}{1+y^2} e^{\tan^{-1}y} dy + C$ $x e^{\tan^{-1}y} = \int \frac{2e^{2\tan^{-1}y}}{1+y^2} dy + C$ Let $t = \tan^{-1}y$, so $\frac{dt}{dy} = \frac{1}{1+y^2}$, and $dt = \frac{1}{1+y^2}dy$. Then, $x e^{\tan^{-1}y} = \int 2e^{2t} dt + C$ $x e^{\tan^{-1}y} = e^{2t} + C$ Substituting back $t = \tan^{-1}y$, we get: $x e^{\tan^{-1}y} = e^{2\tan^{-1}y} + C$ $x = e^{\tan^{-1}y} + Ce^{-\tan^{-1}y}$

Step 4: Apply the Initial Condition to Find the Constant $C$

We are given $f(0) = 1$, which means when $y=0$, $x=1$. Substituting these values: $1 = e^{\tan^{-1}(0)} + Ce^{-\tan^{-1}(0)}$ Since $\tan^{-1}(0) = 0$, we have: $1 = e^0 + Ce^0$ $1 = 1 + C$ $C = 0$

Step 5: Find the Value of $f\left(\frac{1}{\sqrt{3}}\right)$

Since $C=0$, the solution is: $x = e^{\tan^{-1}y}$ We need to find $f\left(\frac{1}{\sqrt{3}}\right)$, so we substitute $y = \frac{1}{\sqrt{3}}$: $x = e^{\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)}$ Since $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$, we have: $x = e^{\pi/6}$

Common Mistakes & Tips

• Remember to correctly identify $P(y)$ and $Q(y)$ after rearranging the differential equation into the standard form.
• Be careful with the integration and substitution steps. A small error can lead to an incorrect solution.
• Don't forget to apply the initial condition to find the constant of integration.

Summary

We solved the given differential equation by first rearranging it into the standard form of a linear first-order differential equation. Then, we found the integrating factor and used it to find the general solution. Finally, we applied the initial condition to find the particular solution and evaluated it at $y = \frac{1}{\sqrt{3}}$ to find the value of $f\left(\frac{1}{\sqrt{3}}\right)$.

The final answer is $\boxed{e^{\pi / 6}}$, which corresponds to option (C).