Find the general solution to the differential equation '' (x^2 + 3x - 1) dy/dx = (2x + 3)y ''

Now although this might seem like quite a complex question at first, it's a little less intimidating when we take a while to look at it- you might notice that we can seperate x and y terms like so:
1/y dy = (2x + 3)/(x^2 + 3x - 1) dx
You're not expected to notice this immediately, but you should eventually get used to seeing when we can rearrange the equation like this, as now we can integrate both sides, the left with respect to y, and the right with respect to x. See, the top of the fraction of the right hand side is the derivative of the bottom- which means the whole thing integrates to log(the bottom of the fraction). You should learn to spot this, it's a useful trick and the examiners tend to like putting it into the papers to catch you off guard.
We now have: '' log y = log ( x^2 + 3x - 1 ) + C '' The constant is important
Put both sides to the power of e to remove the logs:
y = ( e^C )( e^log[x^2 + 3x -1] ) y= A(x^2 + 3x - 1) And we have our general solution! Notice how we've relabeled the constant e^C as ''A''- it makes the equation a lot simpler and is generally allowed, as it is mathematically correct and lets us see how the solution works a little more clearly.