Let y = y(x) be the solution of the differential equation x d y d x - sin 2 y = x 3 ( 2 - x 3 ) cos 2 y , x 0 . If y(2) = 0, then tan ( y ( 1 ) ) is equal to... [2026]

Question

Let y = y(x) be the solution of the differential equation $x \frac{d y}{d x} - \sin 2 y = x^{3} (2 - x^{3}) \cos^{2} y, x \neq 0 .$ If y(2) = 0, then $\tan (y (1))$ is equal to [2026]

Smart Achievers · Accepted Answer

(3)

$x \frac{d y}{d x} - \sin 2 y = x^{3} (2 - x^{3}) \cos^{2} y$

$\sec^{2} y \frac{d y}{d x} - 2 \tan y \cdot \frac{1}{x} = x^{2} (2 - x^{3})$

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

$\frac{d t}{d x} - \frac{2 t}{x} = x^{2} (2 - x^{3}) (LDE)$

$I.F. = e^{\int - \frac{2}{x} d x} = e^{- 2 l n x} = \frac{1}{x^{2}}$

$∴$ $\frac{t}{x^{2}} = \int \frac{1}{x^{2}} x^{2} (2 - x^{3}) d x + C$

$\frac{\tan y}{x^{2}} = 2 x - \frac{x^{4}}{4} + C$

$y (2) = 0 \Rightarrow 0 = 4 - 4 + C \Rightarrow C = 0$

$\tan y = 2 x^{3} - \frac{1}{4} x^{6}$

$x = 1 \Rightarrow \tan y = 2 - \frac{1}{4} = \frac{7}{4}$

Solution

Q.
Let y = y(x) be the solution of the differential equation $x \frac{d y}{d x} - \sin 2 y = x^{3} (2 - x^{3}) \cos^{2} y, x \neq 0 .$ If y(2) = 0, then $\tan (y (1))$ is equal to [2026]

1 $\frac{3}{4}$

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

3 $\frac{7}{4}$

4 $- \frac{7}{4}$

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Solution

Q. Let y = y(x) be the solution of the differential equation xdydx-sin2y=x3(2-x3)cos2y, x≠0. If y(2) = 0, then tan(y(1)) is equal to [2026]

1 34

2 -34

3 74

4 -74