If for the solution curve y = f(x) of the differential equation , , , then is equal to: [2025]

Question

If for the solution curve y = f(x) of the differential equation $\frac{d y}{d x} + (\tan x) y = \frac{2 + \sec x}{{(1 + 2 \sec x)}^{2}}$ , $x \in (\frac{- π}{2}, \frac{π}{2})$ , $f (\frac{π}{3}) = \frac{\sqrt{3}}{10}$ , then $f (\frac{π}{4})$ is equal to: [2025]

Smart Achievers · Accepted Answer

(3)

We have, $\frac{d y}{d x} + (\tan x) y = \frac{2 + \sec x}{{(1 + 2 \sec x)}^{2}}$

Here, P = tan x and $Q = \frac{2 + \sec x}{{(1 + 2 \sec x)}^{2}}$

$I . F . = e^{\int P d x} = e^{\int \tan x d x} = e^{\log | \sec x |} = \sec x$

$∴$ Solution is given by,

$y \cdot \sec x = \int \frac{(2 + \sec x)}{{(1 + 2 \sec x)}^{2}} \cdot \sec x d x + c$

$\Rightarrow y \cdot \sec x = \int \frac{2 \cos x + 1}{{(\cos x + 2)}^{2}} d x + c$

Put $\cos x = \frac{1 - t^{2}}{1 + t^{2}} \Rightarrow - \sin x d x = \frac{- 4 t}{{(1 + t^{2})}^{2}} d t$

$∴ y \cdot \sec x = \int \frac{2 (\frac{1 - t^{2}}{1 + t^{2}}) + 1}{{(\frac{1 - t^{2}}{1 + t^{2}} + 2)}^{2}} \cdot \frac{2 d t}{(1 + t^{2})} + c$

$\Rightarrow y \cdot \sec x = \int \frac{2 - 2 t^{2} + 1 + t^{2}}{{(1 - t^{2} + 2 + 2 t^{2})}^{2}} \cdot 2 d t + c$

$\Rightarrow y \cdot \sec x = 2 \int \frac{3 - t^{2}}{{(t^{2} + 3)}^{2}} d t + c$

Put $t + \frac{3}{t} = u \Rightarrow (1 - \frac{3}{t^{2}}) d t = d u$

$∴ y \cdot \sec x = - 2 \int \frac{d u}{u^{2}} + c$

$\Rightarrow y \cdot \sec x = \frac{2}{u} + c \Rightarrow y \cdot \sec x = \frac{2}{t + 3 / t} + c$

Now, at $x = \frac{π}{3}, t = \frac{1}{\sqrt{3}}, y = \frac{\sqrt{3}}{10}$

$\Rightarrow \frac{2 \sqrt{3}}{10} = \frac{2}{1 / \sqrt{3} + 3 \sqrt{3}} + c \Rightarrow c = 0$

Hence, $y \cdot \sec x = \frac{2}{t + 3 / t} \Rightarrow y = \frac{2}{\sec x \cdot (t + 3 / t)}$

At $x = \frac{π}{4}, t = \sqrt{2} - 1$

$∴ f (\frac{π}{4}) = \frac{1}{\sqrt{2}} \cdot \frac{2 (\sqrt{2} - 1)}{{(\sqrt{2} - 1)}^{2} + 3}$

$= \frac{\sqrt{2} (\sqrt{2} - 1)}{6 - 2 \sqrt{2}} = \frac{2 - \sqrt{2}}{6 - 2 \sqrt{2}} \times \frac{6 + 2 \sqrt{2}}{6 + 2 \sqrt{2}}$

$= \frac{12 - 2 \sqrt{2} - 4}{36 - 8} = \frac{8 - 2 \sqrt{2}}{28} = \frac{4 - \sqrt{2}}{14}$

Solution

Q.
If for the solution curve y = f(x) of the differential equation $\frac{d y}{d x} + (\tan x) y = \frac{2 + \sec x}{{(1 + 2 \sec x)}^{2}}$ , $x \in (\frac{- π}{2}, \frac{π}{2})$ , $f (\frac{π}{3}) = \frac{\sqrt{3}}{10}$ , then $f (\frac{π}{4})$ is equal to: [2025]

1 $\frac{9 \sqrt{3} + 3}{10 (4 + \sqrt{3})}$

2 $\frac{\sqrt{3} + 1}{10 (4 + \sqrt{3})}$

3 $\frac{4 - \sqrt{2}}{14}$

4 $\frac{5 - \sqrt{3}}{2 \sqrt{2}}$

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Solution

Q. If for the solution curve y = f(x) of the differential equation dydx+(tan x)y=2+sec x(1+2 sec x)2, x∈(–π2,π2), f(π3)=310, then f(π4) is equal to: [2025]

1 93+310(4+3)

2 3+110(4+3)

3 4–214