differential equation with substituion

Question

Solve for y:

$y'\tan(x+y)=1-\tan(x+y)$

so far I have made the substituion $u=x+y$, which yields $\frac{du}{dx}=1+\frac{dy}{dx}$. However, I am not sure what to do from here.

Answer 1

Now substitute into the original equation: $y = u - x$, $y' = u' - 1$. You should end up with a separable differential equation.

Answer 2

Hint:

\begin{align*} \frac{dy}{dx}\tan(x+y) &= 1 - \tan(x+y) \ \ \ || \dot\ dx \ \ \ (1) \\\tan(x+y)dy &= 1-tan(x+y)dx \\\int\tan(x+y)dy &= \int 1-tan(x+y)dx \end{align*}

$(1)$ is called Leibniz method (correct me if I'm wrong), and it is quite handy in solving ODE's.

Answer 3

The equation being $$\frac{dy}{dx}\tan(x+y)=1-\tan(x+y)$$ change, as you did $x+y=u$, that is to say $y=u-x$, $\frac{dy}{dx}=\frac{du}{dx}-1$. So, the equation rewrite $$(\frac{du}{dx}-1)\tan(u)=1-\tan(u)$$ that is to say $$\frac{du}{dx}=\cot(u)$$ in other words $$\frac{dx}{du}=\tan(u)$$ Integrating both sides $$x+C=-\log (\cos (u))$$ $$\cos(u)=Ce^{-x}$$ $$u=\cos ^{-1}\left(Ce^{-x}\right)$$ $$y=-x+\cos ^{-1}\left(Ce^{-x}\right)$$