I'm stucked with sovling this equation.
$$y'=\sin(y-x-1)$$
I know I have to use change of variables e.g. $z=y-x-1$ and $z'=y'-1$ but then I am not getting anywhere
$\endgroup$ 32 Answers
$\begingroup$Using the substitution $z = y-x-1$, you get: $$y' = \sin(y-x-1) \Rightarrow z' = \sin(z) - 1$$
This is separable, hence:
$$\frac{dz}{\sin(z) - 1} = dx \Rightarrow \\ \frac{2}{\tan\left(\frac{z}{2}\right) - 1} = x + c \Rightarrow \\ \tan\left(\frac{z}{2}\right) = \frac{2+x+c}{x+c} \Rightarrow \\ z(x) = 2\arctan\left[\frac{2+x+c}{x+c}\right].$$
Then:
$$y(x) = z(x) + x + 1 = x+ 1 + 2\arctan\left[\frac{2+x+c}{x+c}\right]$$
$\endgroup$ 2 $\begingroup$setting $$u=y-x-1$$ then we get $$y'=u'+1$$ and our equation is $$u'+1=\sin(u)$$ solving this equation we get $$u \left( x \right) =2\,\arctan \left( {\frac {{\it \_C1}+x+2}{{\it \_C1}+x}} \right) $$
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