Let \(y=y(x)\) be the solution curve of the differential equation \(\sin \left(2 x^{2}\right) \log _{e}\left(\tan x^{2}\right) d y+\left(4 x y-4 \sqrt{2} x \sin \left(x^{2}-\frac{\pi}{4}\right)\right) d x=0\), \(0<x<\sqrt{\frac{\pi}{2}}\), which passes through the point \(\left(\sqrt{\frac{\pi}{6}}, 1\right)\). Then \(\left|y\left(\sqrt{\frac{\pi}{3}}\right)\right|\) is equal to ______.

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Differential Equations

Let

y = y (x)

be the solution curve of the differential equation

sin (2 x^{2}) lo g_{e} (tan x^{2}) d y + (4 x y - 42 x sin (x^{2} - \frac{π}{4})) d x = 0

0 < x < \frac{π}{2}

, which passes through the point

(\frac{π}{6}, 1)

. Then

y (\frac{π}{3})

is equal to ______.

Answer: 1

Solution

This is a first-order differential equation that can be solved by recognizing it as an exact differential or by finding an integrating factor.

Rewrite the equation:

ln (tan x^{2}) d y + \frac{4 x y}{sin ( 2 x ^{2} )} d x - \frac{4 2 x sin ( x ^{2} - π /4 )}{sin ( 2 x ^{2} )} d x = 0

The key insight is to recognize that this can be written as:

d (y ln (tan x^{2})) + (terms involving x only) = 0

Using the substitution

t = sin x^{2} + cos x^{2}

and simplifying the second term, we get:

y ln (tan x^{2}) + ln (\frac{sin x ^{2} + cos x ^{2} - 1}{sin x ^{2} + cos x ^{2} + 1}) = c

Applying the initial condition

(x, y) = (π /6, 1)

:
At

x = π /6

, we have

x^{2} = π /6

, so

sin x^{2} = 1/2

and

cos x^{2} = 3 /2

.

This gives us the constant

c

.

Now substituting

x = π /3

(so

x^{2} = π /3

sin x^{2} = 3 /2

cos x^{2} = 1/2

y ln (3) + ln (\frac{3 /2 + 1/2 - 1}{3 /2 + 1/2 + 1}) = c

Solving for

y

, we find

y = - 1

.

Therefore,

∣ y (π /3) ∣ = 1

JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Integer answer

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JEE Main · 2022 · 27 Jul 2022 · Shift 1 · Differential Equations

Solution

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