Let y = y(x)$y = y(x)$ be the solution of the differential equation x (dy)/(dx) - 2y = x³(2 - x³) ² y$x \frac{dy}{dx} - \sin 2y = x^{3}(2 - x^{3}) \cos^{2} y$, x ≠ 0$x \neq 0$. If y(2) = 0$y(2) = 0$, then (y(1))$\tan(y(1))$ is equal to
A.(3)/(4)$\frac{3}{4}$
B.(7)/(4)$\frac{7}{4}$
C.-(7)/(4)$-\frac{7}{4}$
D.-(3)/(4)$-\frac{3}{4}$
Solution & Explanation
Core Logic
Linear Differential Equations
Given differential equation:
x (dy)/(dx) - 2y = x³(2 - x³) ² y$$ x \frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y $$
Divide both sides by ² y$\cos^2 y$:
x ² y (dy)/(dx) - (2 y y)/( ² y) = x³(2 - x³)$$ x \sec^2 y \frac{dy}{dx} - \frac{2 \sin y \cos y}{\cos^2 y} = x^3(2 - x^3) $$x ² y (dy)/(dx) - 2 y = x³(2 - x³)$$ x \sec^2 y \frac{dy}{dx} - 2 \tan y = x^3(2 - x^3) $$
Divide by x$x$:
² y (dy)/(dx) - (2)/(x) y = x²(2 - x³)$$ \sec^2 y \frac{dy}{dx} - \frac{2}{x} \tan y = x^2(2 - x^3) $$
Step 1: Reduce to Linear Form
Let y = t$\tan y = t$. Then ² y (dy)/(dx) = (dt)/(dx)$\sec^2 y \frac{dy}{dx} = \frac{dt}{dx}$.
The equation becomes a standard Linear Differential Equation:
(dt)/(dx) - ((2)/(x)) t = x²(2 - x³)$$ \frac{dt}{dx} - \left(\frac{2}{x}\right) t = x^2(2 - x^3) $$
Let y = y(x)$y = y(x)$ be the solution of the differential equation (dy)/(dx) = (( x) + y)/( x( x - x x)), x in (0, (π)/(2))$\frac{dy}{dx} = \frac{(\tan x) + y}{\sin x(\sec x - \sin x \tan x)}, x \in \left(0, \frac{\pi}{2}\right)$ satisfying the condition y((π)/(4)) = 2$y\left(\frac{\pi}{4}\right) = 2$. Then, y((π)/(3))$y\left(\frac{\pi}{3}\right)$ is
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