Key Concepts and Formulas

• Linear First-Order Differential Equation: A differential equation of the form $\frac{dz}{dx} + P(x)z = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$. The integrating factor is given by $e^{\int P(x) dx}$. The solution is $z(x)e^{\int P(x) dx} = \int Q(x)e^{\int P(x) dx} dx + C$.
• Trigonometric Identities: $\frac{\sin y}{\cos y} = \tan y$ and $\frac{1}{\cos y} = \sec y$.
• Integration: $\int x^n dx = \frac{x^{n+1}}{n+1} + C$ for $n \neq -1$ and $\int e^x dx = e^x + C$.

Step-by-Step Solution

Step 1: Transform the differential equation

We are given: $\sec y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x \sin y=x^3 \cos y$ Multiply both sides by $\cos y$: $\frac{\mathrm{d} y}{\mathrm{~d} x} + 2x \sin y \cos y = x^3 \cos^2 y$ Divide both sides by $\cos^2 y$: $\sec^2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+2 x \tan y=x^3$ The reason for doing this is to prepare for a substitution that will linearize the equation.

Step 2: Perform the substitution

Let $z = \tan y$. Then, $\frac{dz}{dx} = \sec^2 y \frac{dy}{dx}$. Substituting into the equation, we get: $\frac{dz}{dx} + 2xz = x^3$ This is now a linear first-order differential equation in $z(x)$.

Step 3: Find the integrating factor

The integrating factor (IF) is given by $e^{\int P(x) dx}$, where $P(x) = 2x$. Thus, $\text{IF} = e^{\int 2x dx} = e^{x^2}$

Step 4: Solve the linear differential equation

Multiply both sides of the linear differential equation by the integrating factor: $e^{x^2} \frac{dz}{dx} + 2xe^{x^2} z = x^3 e^{x^2}$ The left side is the derivative of $z e^{x^2}$ with respect to $x$: $\frac{d}{dx}(z e^{x^2}) = x^3 e^{x^2}$ Integrate both sides with respect to $x$: $\int \frac{d}{dx}(z e^{x^2}) dx = \int x^3 e^{x^2} dx$ $z e^{x^2} = \int x^3 e^{x^2} dx + C$ To evaluate $\int x^3 e^{x^2} dx$, let $u = x^2$, so $du = 2x dx$. Then $x^2 = u$ and $x dx = \frac{1}{2} du$. $\int x^3 e^{x^2} dx = \int x^2 e^{x^2} x dx = \int u e^u \frac{1}{2} du = \frac{1}{2} \int u e^u du$ Using integration by parts, $\int u e^u du = u e^u - \int e^u du = u e^u - e^u + C_1$. So, $\int x^3 e^{x^2} dx = \frac{1}{2} (x^2 e^{x^2} - e^{x^2}) + C_1$ Therefore, $z e^{x^2} = \frac{1}{2} (x^2 e^{x^2} - e^{x^2}) + C$ $z = \frac{1}{2} (x^2 - 1) + Ce^{-x^2}$

Step 5: Substitute back for $z$ and apply the initial condition

Since $z = \tan y$, we have: $\tan y = \frac{1}{2} (x^2 - 1) + Ce^{-x^2}$ We are given the initial condition $y(1) = 0$. Plugging in $x = 1$ and $y = 0$, we get: $\tan 0 = \frac{1}{2} (1^2 - 1) + Ce^{-1^2}$ $0 = 0 + Ce^{-1}$ Thus, $C = 0$.

Step 6: Find the particular solution

The particular solution is: $\tan y = \frac{1}{2} (x^2 - 1)$

Step 7: Evaluate $y(\sqrt{3})$

We want to find $y(\sqrt{3})$. Plugging in $x = \sqrt{3}$, we get: $\tan y(\sqrt{3}) = \frac{1}{2} ((\sqrt{3})^2 - 1) = \frac{1}{2} (3 - 1) = \frac{1}{2} (2) = 1$ Since $\tan y(\sqrt{3}) = 1$, we have $y(\sqrt{3}) = \frac{\pi}{4}$.

Step 8: Identify the correct option

The value of $y(\sqrt{3})$ is $\frac{\pi}{4}$. This corresponds to option (D).

Common Mistakes & Tips

• Remember to use an integrating factor when solving linear first-order differential equations.
• Don't forget the constant of integration when performing indefinite integrals.
• Be careful with the chain rule when performing substitutions.

Summary

We transformed the given non-linear differential equation into a linear first-order differential equation using a suitable substitution. We then found the integrating factor, solved the linear equation, and applied the initial condition to find the particular solution. Finally, we evaluated the solution at $x = \sqrt{3}$ to find $y(\sqrt{3}) = \frac{\pi}{4}$.

The final answer is \boxed{\frac{\pi}{4}}. This corresponds to option (D).