Toronto Math Forum
MAT244--2019F => MAT244--Test & Quizzes => Quiz-2 => Topic started by: lilywq on October 04, 2019, 04:41:26 PM
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\begin{align*}
x^2y^3 + x(1+y^2)y' &= 0 && \equation{\mu = 1/xy^3}\\
1/xy^3*(x^2y^3 + x(1+y^2)y') &= 0 \\
x+y^{-3}(1+y^2)y' &= 0\\
M=x,N=y^{-3}(1+y^2)\\
M_y(x,y)=0=N_x(x,y)\\
\psi_x(x,y)=M(x,y)=x, \psi_y(x,y)=N(x,y)=y^{-3}(1+y^2)\\
\psi=x^2/2+h(y)\\
\psi_y=h'(y)=y^{-3}(1+y^2)\\
h(y)=ln(x)-1/(2y^2)\\
\psi=x^2/2+ln(x)-1/(2y^2)=c\\
\psi=x^2+2ln(x)-y^{-2})=c
\end{align*}