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In answering a differential equation question, the question asked me to solve the equation and then give the function to which $y$ approximates when $x$ is large and positive. To which I have no idea about.

The differential equation was $$\frac{dy}{dx} + \frac{y}{x} = \sin2x$$ given that $y=\frac{2}{\pi}$ when $x=\frac{\pi}{4}$ To which I got the solution $$y=-\frac12 \cos2x + \frac1{4x}\sin2x + \frac1{4x}$$

The question no goes on to say "write down a function to which $y$ approximates when $x$ is large and positive". I have no idea about this, as when i graphed the curve the function quickly seems to just become $-\frac12\cos2x$ and I have no idea why. If someone could explain this to be it would be greatly appreciated.

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  • $\begingroup$ When $x$ is large, both $\frac1{4x}\sin 2x$ and $\frac1{4x}$ terms will be small, close to zero. $\endgroup$ – peterwhy Apr 21 at 19:27
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The reason for this is because $$\frac1{4x}\sin 2x+\frac1{4x}=\frac{1+\sin2x}{4x}.$$ The numerator is always between $0$ and $2$ (inclusive), regardless of $x,$ but the denominator is growing very large, so the fraction becomes very small.

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If the actual $y$ is

$$y(x) = -\frac12\cos 2x + \frac1{4x}\sin 2x +\frac1{4x}$$

And the approximate $y$ is

$$y'(x) = -\frac12\cos 2x$$

Then the difference between the approximation and the actual values of $y$ tends to zero:

$$\lim_{x\to+\infty} [y(x) - y'(x)] = \lim_{x\to+\infty}\left(\frac1{4x}\sin 2x +\frac1{4x}\right) = 0$$

(Though I suppose one could also say that $y$ approximates $-\frac12\cos 2x + \frac1{4x}\sin 2x +\frac1{4x}$)

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