Simple algebra in a differential equation.

Question

I have the differential equation: $$\frac{dy}{dx}=\sin (x-y).$$ Substituting $v=x-y$ and $dy=dx-dv$, I got down to the equation:$$\frac{dv}{1-\sin(v)}=dx.$$ Multiplying the LHS by $\dfrac{1+\sin (v)}{1+\sin(v)}$, I got:$$(\sec^2(v)+\tan (v)\sec (v))dv=dx.$$ This is an easy integral, and I got that $x=\tan(v)+\sec(v)$, with some constant of integration. Now, doing this in Maple gives the result: $$\frac{2}{1-\tan(\frac{v}{2})}=x,$$ where there should be some constant of integration. Now, I need to use "simple" algebra to show that the equation from Maple leads to: $$x=\frac{1+\tan(\frac{v}{2})}{1-\tan(\frac{v}{2})}+C.$$ How do I show that those two equations are equal? I cannot see the simple algebra that needs to be done.

Answer 1

You have $$ x=\frac{2}{1-\tan(\frac{v}{2})}+C.\tag1 $$ Since $C$ just a constant, then you can manipulate it as follows $$ \begin{align} x&=\frac{2}{1-\tan(\frac{v}{2})}+C\\ x&=\frac{2}{1-\tan(\frac{v}{2})}+(K-1)\\ x&=\frac{2}{1-\tan(\frac{v}{2})}-1+K\\ x&=\frac{2-1+\tan(\frac{v}{2})}{1-\tan(\frac{v}{2})}+K\\ x&=\frac{1+\tan(\frac{v}{2})}{1-\tan(\frac{v}{2})}+K.\tag2 \end{align} $$ Thus, $(1)$ and $(2)$ are equal.

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Answer 2

A more computer-algebra-oriented Answer: To show that two expressions differ by a constant, simplify the derivative of their difference. If you get zero, then they differ by a constant. In Maple, for this problem we get

simplify(diff(2/(1-tan(v/2)) - (1+tan(v/2))/(1-tan(v/2)), v));

and Maple responds

0