Question:

Step 1: Identify the type of differential equation. The given differential equation is of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$, which is a first-order linear differential equation. Comparing the given equation $\dfrac{dy}{dx}-\dfrac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}}y=2x\exp\!\left(\dfrac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right)$ with the standard form, we have: $$ P(x) = -\dfrac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}} $$ $$ Q(x) = 2x\exp\!\left(\dfrac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right) $$ (Note: Assuming a likely typo in the question, the denominator in the exponential part of $Q(x)$ is taken as $\sqrt{1+x^6}$ instead of $\sqrt{(1+x)^6}$ for consistency with $P(x)$ and solvability of the problem to match the given answer options.) Step 2: Calculate the integrating factor (I.F.). The integrating factor is given by $I.F. = e^{\int P(x) dx}$. Let's calculate the integral of $P(x)$: $$ \int P(x) dx = \int -\dfrac{3x^5\tan^{-1}(x^3)}{(1+x^6)^{3/2}} dx $$ We use integration by parts, $\int u \, dv = uv - \int v \, du$. Let $u = \tan^{-1}(x^3)$ and $dv = -\dfrac{3x^5}{(1+x^6)^{3/2}} dx$. Then $du = \dfrac{1}{1+(x^3)^2} \cdot 3x^2 dx = \dfrac{3x^2}{1+x^6} dx$. To find $v$, integrate $dv$: Let $w = 1+x^6$, so $dw = 6x^5 dx$. Thus, $3x^5 dx = \frac{1}{2} dw$. $$ v = \int -\dfrac{3x^5}{(1+x^6)^{3/2}} dx = \int -\dfrac{1}{2} w^{-3/2} dw = -\frac{1}{2} \left( -2w^{-1/2} \right) = w^{-1/2} = \dfrac{1}{\sqrt{1+x^6}} $$ Now apply integration by parts: $$ \int P(x) dx = \tan^{-1}(x^3) \cdot \dfrac{1}{\sqrt{1+x^6}} - \int \dfrac{1}{\sqrt{1+x^6}} \cdot \dfrac{3x^2}{1+x^6} dx $$ $$ \int P(x) dx = \dfrac{\tan^{-1}(x^3)}{\sqrt{1+x^6}} - \int \dfrac{3x^2}{(1+x^6)^{3/2}} dx $$ Step 3: Evaluate the remaining integral for the I.F. exponent. Consider the derivative of $\dfrac{x^3}{\sqrt{1+x^6}}$: $$ \dfrac{d}{dx} \left( \dfrac{x^3}{\sqrt{1+x^6}} \right) = \dfrac{3x^2\sqrt{1+x^6} - x^3 \cdot \frac{1}{2\sqrt{1+x^6}} \cdot (6x^5)}{(\sqrt{1+x^6})^2} $$ $$ = \dfrac{3x^2\sqrt{1+x^6} - \dfrac{3x^8}{\sqrt{1+x^6}}}{1+x^6} = \dfrac{3x^2(1+x^6) - 3x^8}{(1+x^6)^{3/2}} = \dfrac{3x^2+3x^8-3x^8}{(1+x^6)^{3/2}} = \dfrac{3x^2}{(1+x^6)^{3/2}} $$ Thus, $\int \dfrac{3x^2}{(1+x^6)^{3/2}} dx = \dfrac{x^3}{\sqrt{1+x^6}}$. Substituting this back into the expression for $\int P(x) dx$: $$ \int P(x) dx = \dfrac{\tan^{-1}(x^3)}{\sqrt{1+x^6}} - \dfrac{x^3}{\sqrt{1+x^6}} = \dfrac{\tan^{-1}(x^3)-x^3}{\sqrt{1+x^6}} $$ Therefore, the integrating factor is: $$ I.F. = \exp\left(\dfrac{\tan^{-1}(x^3)-x^3}{\sqrt{1+x^6}}\right) $$ Step 4: Find the general solution. The general solution is given by $y \cdot I.F. = \int Q(x) \cdot I.F. \, dx + C$. Let's evaluate $Q(x) \cdot I.F.$: $$ Q(x) \cdot I.F. = 2x\exp\!\left(\dfrac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right) \cdot \exp\left(\dfrac{\tan^{-1}(x^3)-x^3}{\sqrt{1+x^6}}\right) $$ Notice that $\exp\!\left(\dfrac{x^3-\tan^{-1}x^3}{\sqrt{1+x^6}}\right) = \exp\!\left(-\left(\dfrac{\tan^{-1}(x^3)-x^3}{\sqrt{1+x^6}}\right)\right) = (I.F.)^{-1}$. So, $Q(x) \cdot I.F. = 2x \cdot (I.F.)^{-1} \cdot I.F. = 2x$. Now, substitute this into the general solution formula: $$ y \cdot \exp\left(\dfrac{\tan^{-1}(x^3)-x^3}{\sqrt{1+x^6}}\right) = \int 2x \, dx + C $$ $$ y \cdot \exp\left(\dfrac{\tan^{-1}(x^3)-x^3}{\sqrt{1+x^6}}\right) = x^2 + C $$ Step 5: Use the initial condition to find the constant $C$. The solution curve passes through the origin, which means $y(0)=0$. Substitute $x=0$ and $y=0$ into the general solution: $$ 0 \cdot \exp\left(\dfrac{\tan^{-1}(0^3)-0^3}{\sqrt{1+0^6}}\right) = 0^2 + C $$ $$ 0 \cdot \exp(0) = 0 + C $$ $$ 0 = C $$ So the particular solution is: $$ y(x) \cdot \exp\left(\dfrac{\tan^{-1}(x^3)-x^3}{\sqrt{1+x^6}}\right) = x^2 $$ $$ y(x) = x^2 \exp\left(-\dfrac{\tan^{-1}(x^3)-x^3}{\sqrt{1+x^6}}\right) $$ $$ y(x) = x^2 \exp\left(\dfrac{x^3-\tan^{-1}(x^3)}{\sqrt{1+x^6}}\right) $$ Step 6: Calculate $y(1)$. Substitute $x=1$ into the particular solution: $$ y(1) = 1^2 \exp\left(\dfrac{1^3-\tan^{-1}(1^3)}{\sqrt{1+1^6}}\right) $$ $$ y(1) = \exp\left(\dfrac{1-\tan^{-1}(1)}{\sqrt{1+1}}\right) $$ Since $\tan^{-1}(1) = \dfrac{\pi}{4}$: $$ y(1) = \exp\left(\dfrac{1-\dfrac{\pi}{4}}{\sqrt{2}}\right) $$ $$ y(1) = \exp\left(\dfrac{\dfrac{4-\pi}{4}}{\sqrt{2}}\right) $$ $$ y(1) = \exp\left(\dfrac{4-\pi}{4\sqrt{2}}\right) $$ Step 7: Conclude the final answer. The value of $y(1)$ is $\exp\!\left(\dfrac{4-\pi}{4\sqrt{2}}\right)$. This matches Option 1. The final answer is $\boxed{\exp\!\left(\dfrac{4-\pi}{4\sqrt{2}}\right)}$.