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.
$\endgroup$4 Answers
$\begingroup$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:
$\endgroup$ $\begingroup$$y'(x) = \frac{c_1}{x}$, and integrating gives $y(x) = c_1 \ln x + c_2$, where $c_2$ is another constant.
We will set $z = y'$ and use the technique of separation of variables.
Your equation reduces to $\displaystyle x \frac{dz}{dx} = -z$, so we arrive at $\displaystyle - \frac{dz}{z} = \frac{dx}{x}$.
Integrating both sides, we get $\log z = - \log x + C$, and $z = \frac{C}{x}$.
Now, substitute $y'$ back in and get $\displaystyle \frac{dy}{dx} = \frac{c_1}{x}$. So $y = c_1 \log x + c_2$.
$\endgroup$ $\begingroup$firts step : u= dy/dx , u' = du/dx
now putting the values of u in main equation : x(du/dx) = -u xdu = -udx du/u = -(1/x) dx taking integral on both sides : lnu = -lnx + c1 to remove ln taking exponential on both sides : elnu = eln-x c1 now we'll left with : u = x^-1 c1 dy/dx = 1/x c1 :. ( u= dy/dx) again taking integral : y = lnxc1 + c2 answer $$xy''+y'=(xy')'=0\to xy'=c\to y'=\dfrac cx\to y=c\log x+d.$$
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