Here's the question you clicked on:
mel0
solve xdy/dx-y=-x
First order linear differential equation. Put it into standard form first by dividing everything by x.
You're first step is to find the integrating factor u(x)
idk if you're there still or not to proceed explaining this...
but your integrating factor is: \[e^{-\ln|x|} = \frac{ 1 }{ x }\] I'm going to skip the work for the next few steps, you can use it as a reference to absolutely check your work, but you SHOULD see that you get product rule as a result. Therefore: \[[\frac{ 1 }{ x }y]'=-1\] Next, take the integral: \[\int\limits [\frac{ 1 }{ x }y]'= - \int\limits dx = \frac{ 1 }{ x }y= -x+C\] \[y = -x^2+Cx\]
the equation should be \[\Large \frac{1}{x}y=-\int\frac{1}{x} (1) dx\]
@sirm3d could you show him how you got that?
the differential equation is \[y'-\frac{1}{x}y=-1\] with integrating factor \(1/x\) the resulting equation is \[(1/x)y=\int(1/x)(-1)dx\]