Key Concepts and Formulas

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

Step-by-Step Solution

Step 1: Identify P(x) and Q(x)

The given differential equation is: $\frac{\mathrm{d} y}{\mathrm{~d} x}-\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} y=2 x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\}$ Comparing this with the general form $\frac{dy}{dx} + P(x)y = Q(x)$, we can identify: $P(x) = -\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}}$ $Q(x) = 2 x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\}$

Step 2: Calculate the Integrating Factor

The integrating factor (IF) is given by $e^{\int P(x) dx}$: $\text{IF} = e^{\int -\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} dx}$ Let $x^3 = t$. Then $3x^2 dx = dt$, and $dx = \frac{dt}{3x^2}$. Also, $x^6 = t^2$, and $x^5 dx = x^3 \cdot x^2 dx = t \cdot \frac{dt}{3} = \frac{t}{3} dt$. Thus, $\int -\frac{3 x^{5} \tan ^{-1}\left(x^{3}\right)}{\left(1+x^{6}\right)^{3 / 2}} dx = \int -\frac{3 \cdot \frac{t}{3} \tan^{-1}(t)}{(1+t^2)^{3/2}} dt = \int -\frac{t \tan^{-1}(t)}{(1+t^2)^{3/2}} dt$ Now, let $u = \tan^{-1}(t)$. Then $t = \tan(u)$ and $dt = \sec^2(u) du$. $\int -\frac{t \tan^{-1}(t)}{(1+t^2)^{3/2}} dt = \int -\frac{\tan(u) \cdot u}{(1+\tan^2(u))^{3/2}} \sec^2(u) du = \int -\frac{\tan(u) \cdot u}{(\sec^2(u))^{3/2}} \sec^2(u) du$ $= \int -\frac{\tan(u) \cdot u}{\sec^3(u)} \sec^2(u) du = \int -\frac{\tan(u) \cdot u}{\sec(u)} du = \int -u \sin(u) du$ Using integration by parts: $\int u dv = uv - \int v du$. Let $u=u$ and $dv = -\sin(u) du$. Then $du = du$ and $v = \cos(u)$. $\int -u \sin(u) du = u \cos(u) - \int \cos(u) du = u \cos(u) - \sin(u) + C_1 = \tan^{-1}(t) \cos(\tan^{-1}(t)) - \sin(\tan^{-1}(t)) + C_1$ Recall that $t = x^3$. Let $\theta = \tan^{-1}(x^3)$. Then $\tan(\theta) = x^3$. We have a right triangle with opposite side $x^3$ and adjacent side $1$. The hypotenuse is $\sqrt{1+x^6}$. Therefore, $\cos(\theta) = \frac{1}{\sqrt{1+x^6}}$ and $\sin(\theta) = \frac{x^3}{\sqrt{1+x^6}}$. $\tan^{-1}(x^3) \cos(\tan^{-1}(x^3)) - \sin(\tan^{-1}(x^3)) = \tan^{-1}(x^3) \frac{1}{\sqrt{1+x^6}} - \frac{x^3}{\sqrt{1+x^6}} = \frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}$ Therefore, the integrating factor is: $\text{IF} = e^{\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}}$

Step 3: Find the General Solution

The general solution is given by: $y \cdot (\text{IF}) = \int Q(x) \cdot (\text{IF}) \, dx + C$ $y \cdot e^{\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}} = \int 2x \exp \left\{\frac{x^{3}-\tan ^{-1} x^{3}}{\sqrt{\left(1+x^{6}\right)}}\right\} e^{\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}} dx + C$ $y \cdot e^{\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}} = \int 2x dx + C$ $y \cdot e^{\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}} = x^2 + C$ $y = (x^2 + C) e^{\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}}$

Step 4: Apply the Initial Condition

The solution curve passes through the origin, so $y(0) = 0$. $0 = (0^2 + C) e^{\frac{0^3 - \tan^{-1}(0^3)}{\sqrt{1+0^6}}} = C e^0 = C$ Therefore, $C = 0$.

Step 5: Find y(1)

The solution is $y = x^2 e^{\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}}$. We want to find $y(1)$. $y(1) = (1)^2 e^{\frac{1^3 - \tan^{-1}(1^3)}{\sqrt{1+1^6}}} = 1 \cdot e^{\frac{1 - \tan^{-1}(1)}{\sqrt{2}}} = e^{\frac{1 - \frac{\pi}{4}}{\sqrt{2}}} = e^{\frac{\frac{4-\pi}{4}}{\sqrt{2}}} = e^{\frac{4-\pi}{4\sqrt{2}}}$

Common Mistakes & Tips

• Sign Errors: Pay close attention to the signs of $P(x)$ and $Q(x)$ when identifying them and calculating the integrating factor.
• Integration by Parts: Remember the formula for integration by parts: $\int u \, dv = uv - \int v \, du$. Choose $u$ and $dv$ carefully to simplify the integral.
• Simplifying Trigonometric Expressions: When using trigonometric substitutions, remember to express your final answer in terms of the original variable.

Summary

We solved the given first-order linear differential equation by finding the integrating factor, applying the initial condition to determine the constant of integration, and then evaluating the solution at $x=1$. The final result is $y(1) = e^{\frac{4-\pi}{4\sqrt{2}}}$.

Final Answer

The final answer is $\boxed{\exp \left(\frac{4-\pi}{4 \sqrt{2}}\right)}$, which corresponds to option (B).