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I have been given the differential equation

$$\frac{dy}{dx}(\tan x) + 2y = x(\operatorname{cosec} x)$$

When $y = 0$ when $x = \pi/2$

Can anyone help me out in this question as I have not done many others like this one?

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Hint: First take $\tan x$ common and then try Linear differential equation of first order

$$\frac{dy}{dx} +P(x)y = Q(x)$$

So here $P(x)=2\cot x$ and $Q(x)=x\dfrac{\cos x}{\sin^2x}$

now just use the standard solution of differential equation

$$ye^{\int P(x)\,dx} = \int Q(x) e^{\int P(x)\,dx} \, dx$$

So you get something like this: $e^{\int P(x)\,dx}=\sin^2x$

$$y\sin^2x = \int x\frac{\cos x}{\sin^2x} (\sin^2x)\,dx$$

$$y\sin^2x = \int x\cos x\,dx$$

Now use Integration by parts rule and then you just satisfy the condition when $y=0$ then $x=\pi /2$ Find the integration constant so you get your unique solution of differential equation..

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    $\begingroup$ Would Q(x) not equal x multiplied by cosx/sin^2x ? $\endgroup$
    – May
    May 2, 2016 at 14:32
  • $\begingroup$ yes you are correct. $\endgroup$ May 2, 2016 at 14:32
  • $\begingroup$ Thank you! Hopefully I can work this out from here! $\endgroup$
    – May
    May 2, 2016 at 14:34
  • $\begingroup$ Proper notation is $\tan x,\ \sin x,\ \cos x$, not $tanx,\ sinx,\ cosx$. If you write a\cos b you see $a\cos b$, but with a cos b you see $a cos b$. The backslash not only de-italicized $\cos$, but also results in proper spacing in things like $a\cos b$ and $a\cos (b)$ (where you'll notice less space to the right of $\cos$ in the second expression; thus spacing depends on context). Also, one conventionally puts a small space between $f(x)$ and $dx$, thus $f(x)\,dx$, coded as f(x)\,dx or f(x)~dx. And you shouldn't continually alternate in and out of MathJax the say you did here. $\endgroup$ Dec 3, 2016 at 17:53

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