Question

(2y ^ 2 + 3x) * d * x + 2xydy = 0 .

Answer 1

$(2y^2+3x)dx+2xydy=0\\ M=2y^2+3xy,~~N=2xy\\ M_y=4y\neq N_x=2y\\ \text{Thus the differential equation is not exact.}\\ \text{But, }\\ \frac{M_y-N_x}{N}=\frac{1}{x} \text{ is a function of x. }\\ I.F= e^{\int\frac{1}{x}dx}=e^{\ln x}=x\\ \text{Multiply the original equation by the integrating factor}\\ (2xy^2+3x^2)dx+2x^2ydy=0\\ M_y=4xy=N_x=4xy\\ \text{Then, it is exact}\\ F_x=M\\ F_x=2xy^²+3x^2\\ \text{Integrate both sides with respect to } x\\ F=x^2y^2+x^3+\Phi(y)\\ \text{Differentiate w.r.t } y\\ F_y=2x^2y+\Phi'(y)\\ 2x^2y=2x^2y+\Phi'(y)\\ \Phi'(y)=0\\ \Phi(y)=C\\ \implies F=x^2y^2+x^3+C$