Question:

Step 1: Identify the type of differential equation. The given differential equation is $\dfrac{dy}{dx}+y\tan x=x\sec x$. This is a linear first-order differential equation of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = \tan x$ and $Q(x) = x\sec x$. Step 2: Calculate the integrating factor (I.F.). The integrating factor is given by $e^{\int P(x) dx}$. $$ \text{I.F.} = e^{\int \tan x dx} $$ The integral of $\tan x$ is $\log|\sec x|$. Since $0 \leq x \leq \dfrac{\pi}{3}$, $\sec x > 0$, so $|\sec x| = \sec x$. $$ \text{I.F.} = e^{\log(\sec x)} = \sec x $$ Step 3: Write the general solution of the differential equation. The general solution is given by $y \cdot (\text{I.F.}) = \int Q(x) \cdot (\text{I.F.}) dx + C$. Substituting the values: $$ y \sec x = \int (x\sec x) (\sec x) dx + C $$ $$ y \sec x = \int x\sec^2 x dx + C $$ Step 4: Evaluate the integral $\int x\sec^2 x dx$ using integration by parts. We use the integration by parts formula $\int u dv = uv - \int v du$. Let $u = x$ and $dv = \sec^2 x dx$. Then $du = dx$ and $v = \int \sec^2 x dx = \tan x$. $$ \int x\sec^2 x dx = x\tan x - \int \tan x dx $$ The integral of $\tan x$ is $\log|\sec x|$. Again, for $0 \leq x \leq \dfrac{\pi}{3}$, $\log|\sec x| = \log(\sec x)$. $$ \int x\sec^2 x dx = x\tan x - \log(\sec x) $$ Step 5: Substitute the evaluated integral back into the general solution for $y$. $$ y \sec x = x\tan x - \log(\sec x) + C $$ Now, divide by $\sec x$ to express $y$ explicitly: $$ y = \frac{x\tan x}{\sec x} - \frac{\log(\sec x)}{\sec x} + \frac{C}{\sec x} $$ Using $\tan x = \dfrac{\sin x}{\cos x}$ and $\sec x = \dfrac{1}{\cos x}$: $$ y = x \frac{\sin x/\cos x}{1/\cos x} - \cos x \log(\sec x) + C\cos x $$ $$ y = x\sin x - \cos x \log(\sec x) + C\cos x $$ Step 6: Use the initial condition $y(0)=1$ to find the constant $C$. Substitute $x=0$ and $y=1$ into the general solution: $$ 1 = (0)\sin(0) - \cos(0)\log(\sec(0)) + C\cos(0) $$ We know $\sin(0)=0$, $\cos(0)=1$, and $\sec(0)=1$. Also, $\log(1)=0$. $$ 1 = 0 - (1)\log(1) + C(1) $$ $$ 1 = 0 - 0 + C $$ $$ C = 1 $$ Step 7: Write the particular solution. Substitute $C=1$ back into the general solution: $$ y = x\sin x - \cos x \log(\sec x) + \cos x $$ Step 8: Evaluate $y\left(\dfrac{\pi}{6}\right)$. Substitute $x=\dfrac{\pi}{6}$ into the particular solution: $$ y\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{6}\sin\left(\dfrac{\pi}{6}\right) - \cos\left(\dfrac{\pi}{6}\right)\log\left(\sec\left(\dfrac{\pi}{6}\right)\right) + \cos\left(\dfrac{\pi}{6}\right) $$ We know the trigonometric values: $\sin\left(\dfrac{\pi}{6}\right) = \dfrac{1}{2}$ $\cos\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{2}$ $\sec\left(\dfrac{\pi}{6}\right) = \dfrac{1}{\cos\left(\dfrac{\pi}{6}\right)} = \dfrac{1}{\sqrt{3}/2} = \dfrac{2}{\sqrt{3}}$ Substitute these values: $$ y\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{6}\left(\dfrac{1}{2}\right) - \dfrac{\sqrt{3}}{2}\log\left(\dfrac{2}{\sqrt{3}}\right) + \dfrac{\sqrt{3}}{2} $$ $$ y\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{12} + \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{2}\log\left(\dfrac{2}{\sqrt{3}}\right) $$ To match the given options, we can combine the $\dfrac{\sqrt{3}}{2}$ term with the logarithm using properties of logarithms: $\log e = 1$. $$ \dfrac{\sqrt{3}}{2} = \dfrac{\sqrt{3}}{2} \log e $$ So, $$ y\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{12} + \dfrac{\sqrt{3}}{2} \log e - \dfrac{\sqrt{3}}{2}\log\left(\dfrac{2}{\sqrt{3}}\right) $$ $$ y\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{12} + \dfrac{\sqrt{3}}{2} \left( \log e - \log\left(\dfrac{2}{\sqrt{3}}\right) \right) $$ Using the property $\log a - \log b = \log\left(\dfrac{a}{b}\right)$: $$ y\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{12} + \dfrac{\sqrt{3}}{2} \log\left(\dfrac{e}{2/\sqrt{3}}\right) $$ $$ y\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{12} + \dfrac{\sqrt{3}}{2} \log\left(\dfrac{e\sqrt{3}}{2}\right) $$ This can also be written as: $$ y\left(\dfrac{\pi}{6}\right) = \dfrac{\pi}{12} - \dfrac{\sqrt{3}}{2} \log\left(\dfrac{2}{e\sqrt{3}}\right) $$ (by changing the sign and inverting the argument of the logarithm). Step 9: State the final answer. The value of $y\left(\dfrac{\pi}{6}\right)$ is $\dfrac{\pi}{12} - \dfrac{\sqrt{3}}{2}\log_e\!\left(\dfrac{2}{e\sqrt{3}}\right)$. This matches Option 1. The final answer is $\boxed{\text{Option 1}}$.