$\displaystyle xy'-sin(x)=0$,$\displaystyle y = \int_{0}^{x}\frac{sin(t)}{T}dt$So this is what I did. I integrated the function to get, $\displaystyle \frac{1}{T}[-cos(x)+cos(1)]$

Then took the derivative of that with respect to x to get $\displaystyle \frac{sin(x)}{T}$

Then I sub it in to get $\displaystyle \frac{xsin(x)}{T} - sin(x)$ which does not equal 0, so i'm saying that this function is not a solution to the differential equation.

IS this correct?

David.