Question

Q #1 (2 marks) Find the general solution for the DExy'-ymx1y, x>0, y>0.

Answer 1

Given ODE xy^1 - y = x^2/y x > 0, y > 0 multiply this equation with y/x on both sides y/x(xy^' - y) = y/x .x^2/y y y^' - y^2/x = x/1 y. y^' - 1/x.y^2 = x ___1 is a Bernoulli equation put y^2 = u implies d/dx(y^2) = du/dx implies 2y dy/dx = du/dx implies y dy/dx = 1/2 du/dx implies y. y^' = 1/2 u^' then circular 1 becomes 1/2 u^' - 1/x.u = x implies 2[1/2 u^' - 1/x u] = 2x implies u^' + (-2/x). u = 2x its in the form u^1 p(x) u = Q(x), where p(x) = -2/x, Q(x) = 2x its integration factor mu = e^integral p(x) dx = e^integral -2/x dx = e^-2 integral 1/x dx = e^2 ln (x) = e^ln (x) = e^ln x^-2 = x^-2 implies mu = 1/x^2 its solution is u(mu) = integral mu. Q(x) dx u(1/x^2) = integral1/x^2 2x dx y^2.1/x^2 = 2 integral 1/x dx (because u = y^2)