Question

Let $y (x)$ be the solution of the differential equation
$\frac{d y}{d x} = 1 + y sec x, for x \in (- \frac{π}{2}, \frac{π}{2})$
that satisfies $y (0) = 0$ . Then, the value of $y (\frac{π}{6})$ equals

$\sqrt{3} log (\frac{3}{2})$
$(\frac{\sqrt{3}}{2}) log (\frac{3}{2})$
$(\frac{\sqrt{3}}{2}) log 3$
$\sqrt{3} log 3$

Answer: $\sqrt{3} log (\frac{3}{2})$

Solution :-

We have,

\begin{aligned} \frac{d y}{d x} & = & 1 + y sec x \\ ⟹ \frac{d y}{d x} - y sec x & = & 1 \end{aligned}

This is a 1st order 1st degree linear differential equation.
Hence, we solve this differential equation by the method of 1st order 1st degree linear differential equation.

So, $\frac{d y}{d x} + (- sec x) y = 1$

\begin{aligned} I.F. & = & e^{\int (- sec x) d x} \\ = & e^{- log (sec x + tan x)} \\ = & e^{log {(sec x + tan x)}^{- 1}} \\ = & \frac{1}{sec x + tan x} \\ = & \frac{cos x}{1 + sin x} \end{aligned}

So, the solution is

\begin{aligned} y \cdot \frac{cos x}{1 + sin x} & = & \int 1 \cdot \frac{cos x}{1 + sin x} d x \\ y \cdot \frac{cos x}{1 + sin x} & = & \int \frac{cos x}{1 + sin x} d x \dots \dots (1) \end{aligned}

Now, let $1 + sin x = t$ , then differentiating w. r. to $x$ , we get

\begin{aligned} \frac{d (1 + sin x)}{d x} & = & \frac{d t}{d x} \\ cos x & = & \frac{d t}{d x} \\ cos x d x & = & d t \end{aligned}

Now, put $cos x d x = d t and 1 + sin x = t$ in eq(1), we get

\begin{aligned} y \cdot \frac{cos x}{1 + sin x} & = & \int \frac{1}{t} d t \\ y \cdot \frac{cos x}{1 + sin x} & = & log t + c \\ y \cdot \frac{cos x}{1 + sin x} & = & log (1 + sin x) + c \dots \dots (2) (∵ 1 + sin x = t) \end{aligned}

Now, given that $y (0) = 0$ , So from eq(2), we have

\begin{aligned} 0 \cdot \frac{cos 0}{1 + sin 0} & = & log (1 + sin 0) + c \\ 0 & = & log (1 + 0) + c \\ 0 & = & 0 + c \\ ∴ c & = & 0 \end{aligned}

Put c = 0 in eq(2), we get

\begin{aligned} y \cdot \frac{cos x}{1 + sin x} & = & log (1 + sin x) + 0 \\ ∴ y & = & \frac{1 + sin x}{cos x} log (1 + sin x) \\ Now, y (\frac{π}{6}) & = & \frac{1 + sin \frac{π}{6}}{cos \frac{π}{6}} log (1 + sin \frac{π}{6}) \\ = & \frac{1 + \frac{1}{2}}{\frac{\sqrt{3}}{2}} log (1 + \frac{1}{2}) \\ = & \frac{\frac{3}{2}}{\frac{\sqrt{3}}{2}} log (\frac{3}{2}) \\ = & \sqrt{3} log (\frac{3}{2}) \cdot \end{aligned}

So, correct answer is option-1, $\to \sqrt{3} log (\frac{3}{2})$

#Ordinary Differential Equation #IIT JAM PYQ 2024 #Mathematics #IIT JAM PYQs

Question

Let y⁢(x) be the solution of the differential equation d⁢yd⁢x=1+ysecx,forx∈(−π2,π2) that satisfies y⁢(0)=0 . Then, the value of y⁢(π6) equals

Solution :-

Let $y (x)$ be the solution of the differential equation
$\frac{d y}{d x} = 1 + y sec x, for x \in (- \frac{π}{2}, \frac{π}{2})$
that satisfies $y (0) = 0$ . Then, the value of $y (\frac{π}{6})$ equals