I have an ODE $\frac{dy}{dx} + y = \sin 2x$. Now I solved the ode where
$$y=Ae^{-x} + \frac{1}{5} \sin 2x - \frac{2}{5} \cos 2x.$$
The question wants me to use the solution to find u(x) for this ODE:
$$\frac{d^2u}{dx^2} + \frac{du}{dx} = \sin 2x.$$ Apparently if I substitute $du/dx$ as $y$ then I integrate both sides I get $u(x)$ .But why is $du/dx=y$ in this second ODE? Clearly the first ODE is a first order differentiation and the second is a second order? Any help will be appreciated!
You are making use of what you learned from the first equation to solve the second. You can simply define $\frac {du}{dx}=y$ and then integrate the solution to the first equation to get an equation for $u$.