# Thread: differential equation with substitution!!!

1. ## differential equation with substitution!!!

using the substitution u = y-x solve dy/dx = sin(y-x) after substitution i get sin u but then how do integrate with respect to x????

2. First thing to notice is that if $\displaystyle u = y - x$, then $\displaystyle \frac{du}{dx} = \frac{dy}{dx} -1$.

Substituting into your equation, then, will give you:

$\displaystyle \frac{du}{dx} = sin(u) +1$.

Do you see where to take it from here?

3. i got to the form int (1/(sin u + 1)) du = int dx but how do integrate the part on the left?

4. Multiply top and bottom by $\displaystyle 1-\sin u.$

5. Originally Posted by Henderson
First thing to notice is that if $\displaystyle u = y - x$, then $\displaystyle \frac{du}{dx} = \frac{dy}{dx} -1$.

Substituting into your equation, then, will give you:

$\displaystyle \frac{du}{dx} = sin(u) +1$.

Do you see where to take it from here?
i tried this but shouldn't the correct form be du/dx = sin (u) - 1 and not plus 1 like you wrote, Henderson??