# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-2 => Topic started by: Kunpeng Liu on October 04, 2019, 03:03:04 PM

Title: TUT 0702 QUIZ2
Post by: Kunpeng Liu on October 04, 2019, 03:03:04 PM
$$Question:Show\, \, that\, \, the\, \, given\, \, equation\, \, is\, \, not\, \, exact\, \, but\, \, becomes \, \, exact\, \, when\, \, multiplied\, \, by\, \, the \, \, given\, \, integrating\, \, factor.\, \, Then\, \, solve \, \,the \, \, equation.\\x^{2}y^{3}+x(1+y^{2}){y}'=0\, \, \, \, \, \, \mu (x,y)=ye^x\\\\Let\, \, M=x^2y^3,\, \, N=x(1+y^2)\\\\My=3x^2y^2,\, \, Nx=1+2xy\, \, \, \, \because My\neq Nx\, \, \, \therefore Not\, \, \, exact\\\\Both\, \, sides\, \, multiplied\, \, \mu :\\\\x+(\frac{1}{y^3}+\frac{1}{y}){y}'=0\, \, \, \, \, \, Let\, \, M=x\, \, \, \, N=\frac{1}{y^3}+\frac{1}{y}\\\\\because My=Nx=0\, \, \therefore Exact\\\\\phi x=\int xdx=\frac{1}{2}x^2+h(y)\, \, \, \, \, \, \, \, \, \, \phi y={h}'(y)=y^{-3}+y^{-1}\\\\h(y)=-\frac{1}{2}y^{-2}-ln\left | y \right |\\\\\therefore \frac{1}{2}x^{2}-\frac{1}{2}y^{-2}-ln\left | y \right |=C$$
Title: Re: TUT 0702 QUIZ2
Post by: Yu Qi Huang on October 04, 2019, 05:56:01 PM
Hey there,

I believe you made a mistake there in your solution as it should be 1/2·x^2 − 1/2·y^(−2) + ln|y| = c
rather than 1/2·x^2 − 1/2·y^(−2) - ln|y| = c
since the derivative of y^(-1) is ln|y| not -ln|y|.