Question

5. Find the general solution to the ODE (2+1) y=x2 + 2x-1-4xy dx

Answer 1

(x^2 + 1) dy/dx = x^2 + 2x - 1 - 4xy (x^2 + 1) dy - (x^2 + 2x - 1 - 4xy) dx = 0 M = -(x^2 + 2x - 1 - 4xy) = x^2 - 2x + 1 + 4xy N = x^2 + 1 partial differential M/partial differential y = 4n partial differential N/partial differential x = 2x f (x) = 1/N (partial differential M/partial differential y - partial differential N/partial differential x) = 1/x^2 + 1 2n Integrating factor I.F = e^integral f(x) dx = e^integral 2x/x^2 + 1 dx = e^integral^dt/t = e^ln t = e^ln (x^2 + 1) = x^2 + 1 x^2 + 1 = t 2x dx = dt So solution of (1) is integral keeping y as constant M dx + integral N dy downarrow not involving x term = C integral -(x^2 - 2x + 1 + 4xy) dx + integral 0 dy = C integral (-x^2 + 2x - 1 - 4xy) dx = C -x^3/3 + 2 x^2/2 - x - 4 y x^2/2 = C -x^3/3 + x^2 - x - 2y x^2 = C is the required solution.