Hello, new here. Really need help with this one nonlinear differential equation problem from my Calc IV homework. Any help or hints would be greatly appreciated!
xy' - y = 1/(x^3)
Thank you in advance!
An integrating factor, for a linear differential equation, is a function, , such that multiplying by it makes the left side an "exact derivative". That is, . Theoretically, all first order differential equations have "integrating factors" but it is only linear equations for which it is easy to find because , by the product rule, is so we must have , a first order, separable, differential equation for . We can separate that as so that , giving the exponential formula Prove It gave.
In this case, so and so that equation is just which is the same as so that we have . That's easy to integrate: . Since we only need one integrating factor, we can take C= 0 and then take the exponential of both sides to get as integrating factor.
Thanks, Prove It and HallsofIvy! Those equations really helped. I wish my textbook was as thorough as that post was.
I worked the integrating factor out and got x^-2, multiplied both sides of the equation and the general solution came out to be -3(x^-2) + Cx.