# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-2 => Topic started by: Jingjing Cui on October 04, 2019, 02:00:35 PM

Title: TUT0402 Quiz2
Post by: Jingjing Cui on October 04, 2019, 02:00:35 PM
$$\\ \frac{x}{({x^2+y^2})^{\frac{3}{2}}}+\frac{y}{({x^2+y^2})^{\frac{3}{2}}}y'=0 \\ M=\frac{x}{({x^2+y^2})^{\frac{3}{2}}} \\ N=\frac{y}{({x^2+y^2})^{\frac{3}{2}}} \\ My=\frac{d}{dy}\frac{x}{({x^2+y^2})^{\frac{3}{2}}}=-2xy\frac{3}{2}({x^2+y^2})^{-\frac{5}{2}}=-\frac{3xy}{({x^2+y^2})^{\frac{5}{2}}} \\ Nx=\frac{d}{dx}\frac{y}{({x^2+y^2})^{\frac{3}{2}}}=-2xy\frac{3}{2}({x^2+y^2})^{-\frac{5}{2}}=-\frac{3xy}{({x^2+y^2})^{\frac{5}{2}}} \\ Therefore\; ,\; it's\; exact\; \\ \\ \phi=\int{M}dx=\int\frac{x}{({x^2+y^2})^{\frac{3}{2}}}dx \\ Let\; u=x^2+y^2\; ,\; then\; du=2xdx\; ,\; xdx=\frac{1}{2}du \\ \phi=\frac{1}{2}\int\frac{1}{u^{\frac{3}{2}}}du=-2\frac{1}{2}\frac{1}{u^{\frac{1}{2}}}+h(y)=-\frac{1}{({x^2+y^2})^{\frac{1}{2}}}+h(y) \\ \\ \phi{y}=\frac{1}{2}2y\frac{1}{({x^2+y^2})^{\frac{3}{2}}}+h'(y)=\frac{y}{({x^2+y^2})^{\frac{3}{2}}}+h'(y)=N \\ Therefore\; ,\; h'(y)=0\; h(y)=k \\ \\ \phi=-\frac{1}{({x^2+y^2})^{\frac{1}{2}}}+k \\ k=-\frac{1}{({x^2+y^2})^{\frac{1}{2}}} \\ ({x^2+y^2})^{\frac{1}{2}}=\frac{-1}{k} \\ {x^2+y^2}=\frac{1}{k^2} \\ Thus\; ,\; the\; solution\; is\; {x^2+y^2}=C\; where\; C=\frac{1}{k^2}$$