Question

Find the solution of the equation that satisfies the given conditions for x-> infinite. The equation is: (x^2)(y') - cos(2y) = 1, y(+infinte) = (9)(pie)/(4)

Answer 1

Answer on Question #34550 – Math – Differential Equations

Find the solution of the equation that satisfies the given conditions for $x$-$>$ infinity. The equation is $(x^2)(y') - \cos(2y) = 1$, $y(+\text{infinity}) = (9)(\text{pie})/(4)$

Solution

$$\begin{aligned} x^2 y' - \cos(2y) &= 1 \rightarrow \frac{dy}{1 + \cos(2y)} = \frac{dx}{x^2} \rightarrow \\ &\rightarrow \int \frac{dy}{1 + \cos(2y)} = \int \frac{dx}{x^2} \rightarrow \frac{\tan(y)}{2} = -\frac{1}{x} + c \rightarrow \\ &\rightarrow \tan(y) = -\frac{2}{x} + 2c \\ &\rightarrow y = \tan^{-1}\left(\frac{2cx - 2}{x}\right) \end{aligned}$$

If $y(\infty) = \tan^{-1}(2c) = \frac{9\pi}{4}$, then

$$\tan(y(\infty)) = \tan\left(\tan^{-1}(2c)\right) = 2c = \tan\left(\frac{9\pi}{4}\right) = 1$$ $$\rightarrow 2c = 1 \rightarrow c = \frac{1}{2}$$

Thus, $y = \tan^{-1}\left(\frac{x - 2}{x}\right)$.

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