When solving ODE by substitution, where does $dy$ goes from the following example? $$\left(1+\frac{sin(y)}{cos(y)}\right)dy=x dx$$ Let $u=-cos(y)$. Hence $du = sin(y)$, which results in the following: $$\left(1-\frac{1}{u}\right)du=xdx$$ But intuitively I want to write: $$\left(\left(1-(\frac{1}{u}du\right)\right)dy=xdx$$ So where did $dy$ go, could someone point out what logic eliminates $dy$ please?
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$\begingroup$ Edits and improvements are welcome. $\endgroup$– Jek DenysCommented Jul 16, 2016 at 4:38
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$\begingroup$ @Moo, Yes. Thank you for pointing that out $\endgroup$– Jek DenysCommented Jul 16, 2016 at 4:56
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The substitution should be $du \equiv \sin{(y)}dy$.
We define $u = -\cos{(y)}$, and so $$\frac{du}{dy} = \sin{(y)}.$$ By abuse of notation, we can intuitively "multiply this fraction" by $dy$— which would intuitively make $du \equiv \sin{(y)}dy$.
The equation $\left(1 + \frac{\sin{y}}{\cos{y}}\right) dy = x \,dx $ becomes
$$\left(1 + \frac{du}{-u}\right) = x \,dx$$
or just $$\left(1-\frac{1}{u}\right)du = x \,dx.$$
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$\begingroup$ O-o-o-h-h... That's what I was missing! $$du = sin(y)du$$ $$\frac{du}{dy}=sin(y)$$ Thank you! $\endgroup$ Commented Jul 16, 2016 at 5:00