Let g be differentiable function such that and let y = y(x) satisfy the differential equation . If y(0) = 0, then is equal to [2025]

Question

Let g be differentiable function such that $\int_{0}^{x} g (t) d t = x - \int_{0}^{x} t g (t) d t, x \geq 0$ and let y = y(x) satisfy the differential equation $\frac{d y}{d x} - y \tan x = 2 (x + 1) \sec x g (x), x \in [0, \frac{π}{2})$ . If y(0) = 0, then $y (\frac{π}{3})$ is equal to [2025]

Smart Achievers · Accepted Answer

(4)

we have, $\int_{0}^{x} g (t) d t = x - \int_{0}^{x} t g (t) d t$ ... (i)

On differentiating equation (i), we get g(x) = 1 – xg(x)

$\Rightarrow g (x) = \frac{1}{1 + x}$

Now, $\frac{d y}{d x} - y \tan x = 2 (x + 1) \sec x g (x)$

$\Rightarrow \frac{d y}{d x} - y \tan x = 2 \sec x$

$∴ I . F . = e^{- \int \tan x d x} = e^{\log_{e} \cos x} = \cos x$

Solution of D.E. is given by

$y \cos x = \int 2 \cos x \cdot \sec x d x + C$

$\Rightarrow y \cos x = 2 x + C \Rightarrow y (0) = 0 \Rightarrow C = 0$

$∴ y = \frac{2 x}{\cos x}$

$\Rightarrow y = 2 x \sec x \Rightarrow y (\frac{π}{3}) = 2 \cdot \frac{π}{3} \cdot 2 = \frac{4 π}{3}$

Solution

Q.
Let g be differentiable function such that $\int_{0}^{x} g (t) d t = x - \int_{0}^{x} t g (t) d t, x \geq 0$ and let y = y(x) satisfy the differential equation $\frac{d y}{d x} - y \tan x = 2 (x + 1) \sec x g (x), x \in [0, \frac{π}{2})$ . If y(0) = 0, then $y (\frac{π}{3})$ is equal to [2025]

1 $\frac{2 π}{3 \sqrt{3}}$

2 $\frac{2 π}{3}$

3 $\frac{4 π}{3 \sqrt{3}}$

4 $\frac{4 π}{3}$

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Solution

Q. Let g be differentiable function such that ∫0xg(t)dt=x–∫0xtg(t)dt, x≥0 and let y = y(x) satisfy the differential equation dydx–y tan x=2(x+1) sec xg(x), x∈[0,π2). If y(0) = 0, then y(π3) is equal to [2025]

1 2π33

2 2π3

3 4π33