Solve the differential equation: $xy'' + y' = 0$ Title of the question states it. I need to solve $xy'' + y' = 0$. Any pointers? I know it's easy but I struggle with differential equations.... Thanks in advance.
 A: Hint: $\frac{d}{dx} (x y'(x)) = x y''(x) + y'(x)$. Since the derivative is zero, the quantity in parentheses is a constant.
This gives:

 $x y'(x) = c_1$, where $c_1$ is a constant.

This can be rewritten as:

 $y'(x) = \frac{c_1}{x}$, and integrating gives $y(x) = c_1 \ln x + c_2$, where $c_2$ is another constant.

A: We will set $z = y'$ and use the technique of separation of variables.
Your equation reduces to $\displaystyle x \frac{dz}{dx} = -z$, so we arrive at $\displaystyle - \frac{dz}{z} = \frac{dx}{x}$.
Integrating both sides, we get $\log z = - \log x + C$, and $z = \frac{C}{x}$.
Now, substitute $y'$ back in and get $\displaystyle \frac{dy}{dx} = \frac{c_1}{x}$.  So $y = c_1 \log x + c_2$.
A: firts step :
             u=  dy/dx   ,   u' = du/dx
        now putting the values of u in main equation :


          x(du/dx) = -u

          xdu = -udx 

          du/u = -(1/x) dx 

          taking integral on both sides : 



           lnu = -lnx + c1

           to remove ln taking exponential on both sides :

           elnu =  eln-x c1

           now we'll left with :

           u =  x^-1 c1
           dy/dx = 1/x c1             :. ( u= dy/dx)
            again taking integral :


           y = lnxc1 + c2      answer 

A: $$xy''+y'=(xy')'=0\to xy'=c\to y'=\dfrac cx\to y=c\log x+d.$$
