Author Topic: TUT0402 Quiz2  (Read 291 times)

Jingjing Cui

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