step 3 requires that , which is a separable differential equation, whose solution yields in terms of : To verify, multiplying by gives By applying the product rule in reverse, we see that the left-hand side can be expressed as a single derivative in We use this fact to simplify our expression to Integrating both sides with respect to where is a constant. Moving the exponential to the right-hand side, the general solution to Ordinary Differential Equation is: In the case of a homogeneous differential equation, and the general solution to Ordinary Differential Equation is: . for example, consider the differential equation We can see that in this case Multiplying both sides by we obtain The above equation can be rewritten as By integrating both sides with respect to x we obtain or The same result may be achieved using the following approach Reversing the quotient rule gives or or where is a constant. Solving second order linear ordinary differential equations The method of integrating factors for first order equations can be naturally extended to second order equations as well. The main goal in solving first order equations was to find an integrating factor such that multiplying by it would yield , after which subsequent integration and division by would yield . For second order linear differential equations, if we want to work as an integrating factor, then This implies that a second order equation must be exactly in the form for the integrating factor to be usable. Example 1 For example, the differential equation can be solved exactly with integrating factors. The appropriate can be deduced by examining the term. In this case, , so . After examining the term, we see that we do in fact have , so we will multiply all terms by the integrating factor . This gives us which can be rearranged to give Integrating twice yields Dividing by the integrating factor gives: Example 2 A slightly less obvious application of second order integrating factors involves the following differential equation: At first glance, this is clearly not in the form needed for second order integrating factors. We have a term in front of but no in front of . However, and from the Pythagorean identity relating cotangent and cosecant, so we actually do have the required term in front of and can use integrating factors. Multiplying each term by gives which rearranged is Integrating