\(\exp\left(\frac{\pi - 4}{4\sqrt{2}}\right)\)
\(\exp\left(\frac{4 - \pi}{4\sqrt{2}}\right) \)
\(\exp\left(\frac{1 - \pi}{4\sqrt{2}}\right)\)
\(\exp\left(\frac{4 + \pi}{4\sqrt{2}}\right)\)
The given differential equation is:
\[ \frac{dy}{dx} - \frac{3x^5 \tan^{-1}(x^3)}{(1 + x^6)^{3/2}} y = 2x. \]
The integrating factor (I.F.) is given by:
\[ \text{I.F.} = e^{\int -\frac{3x^5 \tan^{-1}(x^3)}{(1 + x^6)^{3/2}} dx}. \]
The integral simplifies as:
\[ \int -\frac{3x^5 \tan^{-1}(x^3)}{(1 + x^6)^{3/2}} dx = \tan^{-1}(x^3) \cdot \frac{x^3}{\sqrt{1 + x^6}}. \]
Thus, the integrating factor becomes:
\[ \text{I.F.} = e^{\tan^{-1}(x^3) \cdot \frac{x^3}{\sqrt{1 + x^6}}}. \]
The general solution of the differential equation is:
\[ y \cdot \text{I.F.} = \int 2x \cdot \text{I.F.} dx + C. \]
Substitute the I.F.:
\[ y \cdot e^{\tan^{-1}(x^3) \cdot \frac{x^3}{\sqrt{1 + x^6}}} = \int 2x dx + C. \]
Simplify:
\[ y \cdot e^{\tan^{-1}(x^3) \cdot \frac{x^3}{\sqrt{1 + x^6}}} = x^2 + C. \]
At \(x = 0\), \(y = 0\). Substitute:
\[ 0 \cdot e^{\tan^{-1}(0) \cdot 0 \cdot \sqrt{1 + 0^6}} = 0^2 + C \implies C = 0. \]
Thus, the solution becomes:
\[ y \cdot e^{\tan^{-1}(x^3) \cdot \frac{x^3}{\sqrt{1 + x^6}}} = x^2. \]
At \(x = 1\):
\[ y(1) \cdot e^{\tan^{-1}(1^3) \cdot \frac{1^3}{\sqrt{1 + 1^6}}} = 1^2. \]
Simplify the terms:
\[ y(1) \cdot e^{\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}} = 1. \]
Rewriting the exponent:
\[ y(1) = e^{-\frac{\pi}{4\sqrt{2}}}. \]
Express the exponent further:
\[ y(1) = \exp\left(\frac{4 - \pi}{4\sqrt{2}}\right). \]
The value of \(y(1)\) is:
\[ \boxed{\exp\left(\frac{4 - \pi}{4\sqrt{2}}\right)}. \]
Therefore, the correct answer is (B).
In the given figure, the blocks $A$, $B$ and $C$ weigh $4\,\text{kg}$, $6\,\text{kg}$ and $8\,\text{kg}$ respectively. The coefficient of sliding friction between any two surfaces is $0.5$. The force $\vec{F}$ required to slide the block $C$ with constant speed is ___ N.
(Given: $g = 10\,\text{m s}^{-2}$) 
Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
