Show that the function is a solution of the differential equation y = xlnx - 4x Differential equation: x + y - xy' = 0

Question

anonymous · Answer

compute the derivative of $x\ln(x)-4x$, replace $y'$ by that derivative, and you will see that you get 0

kirbykirby · Answer

Find the derivative of your function first (to find y'):
y' = (1)lnx + x(1/x) - 4 = lnx+1-4 = lnx - 3

Now you substitute it into the differential equation:

x + (xlnx - 4x) -x(lnx - 3) 
= x + xlnx - 4x -xlnx + 3x
= x - x = 0

anonymous · Answer

$$y'=1+\ln(x)-4$$ now 
$$x+y-x(1+\ln(x)-4)=0$$ is true

anonymous · Answer

Thank you so much guys! My teacher assigned tons of problems this morning and I have been staring at the stuff all day so this is really helping me out. :)