First order differential equations are equations which include a first derivative ie) dy/dx if the equation is in terms of x and y.

Solving these equations involves finding a expression for one of the variables in terms of the other without including a derivative.

To do this usually a general solution is found, one which includes a constant from integration, and then using a given set of conditions (usually initial conditions) the particular solution can be found.

This is done by “splitting the variables” and then integrating the resulting equation. Splitting the variable means that we treat as a fraction so we can “split” the top from the bottom to take the bottom to the other side. What we want is for all the x terms to be on one side and all the y terms on the other. We can then “cross multiply” so that all the x terms are “multiplied” by dx and all the y terms are “multiplied” by dy. We can then add in the integrals and integrate each side (one with respect to x and the other with respect to y).

[Of course this isn't what is actually happening as isn't a fraction and so we can't multiply by dx. What happens in reality is that we can rearrange so that all of the y's are on the side of the and all of the x's are on the other. Now we have a function of y multiplied by the derivative of y with respect to x. Now we can integrate both sides with respect to x. The side with the x's using normal rules of integration but on the side with the y we use a change of variable from x to y, to do this we must multiply by or (these can be proven to be the same) which will cancel out the and hence we can simply integrate the expression on the side with the y's with respect to y giving us the same result as when we "multiply".]

Then simply re arranging will give the general solution (that is you can rearrange it so that the equation is of the form y=…+C). If you need the particular solution substitute the initial conditions into the equation to find out the value of the constant and then substitute this value for the constant into the general solution.

If all of that was a bit hard to follow here is s a worked example.

Example

Find the general solution to
dy/dx = 3x + 4
and the particular solution in the case y = 4 when x = 0 .
So we can rearrange to get

dy = (3x+4)dx

And then by integrating both sides we get

∫dy = ∫3x+4 dx

y = (3x+4)²/6 + c

Which is the general solution.
We can use boundary conditions, ie) y = 4 when x = 0 to find a particular solution
4 = (0+4)²/6 +c
c = 4-16/6 = 8/6 = 4/3

so the particular solution is
y = (3x+4)²/6 +4/3