Question
Let $y = y(x)$ be the solution of the differential equation
$x\frac{dy}{dx} - sin2y = x^{3}\left( 2 - x^{3} \right)\cos^{2}y,x \neq 0$.
If $y(2) = x$, then $tan(y(1))$ is equal to
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Solution
$x\frac{dy}{dx} - sin2y = x^{3}\left( 2 - x^{3} \right)\cos^{2}y$
$${\sec^{2}y\frac{dy}{dx} - 2tany \cdot \frac{1}{x} = x^{2}\left( 2 - x^{3} \right) }{tany = t \Rightarrow \sec^{2}y\frac{dy}{dx} = \frac{dt}{dx} }$$$\frac{dt}{dx} - \frac{2t}{x} = x^{2}\left( 2 - x^{3} \right)$ (LDE)
I.F. $= e^{\int - \frac{2}{x}dx} = e^{- 2lnx} = \frac{1}{x^{2}}$
$${\therefore\frac{t}{x^{2}} = \int\frac{1}{x^{2}}x^{2}\left( 2 - x^{3} \right)dx + C }{\frac{tany}{x^{2}} = 2x - \frac{x^{4}}{4} + C }{y(2) = 0 \Rightarrow 0 = 4 - 4 + C \Rightarrow C = 0 }{tany = 2x^{3} - \frac{1}{4}x^{6}}$$
$$\begin{array}{r} x = 1 \Rightarrow tany = 2 - \frac{1}{4} = \frac{7}{4} \Rightarrow \#(2) \end{array}$$
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