Author Topic: TUT0302 Quiz2  (Read 309 times)

Aiting Zhang

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TUT0302 Quiz2
« on: October 04, 2019, 02:00:01 PM »
$$\mbox{Solve the equation }$$
$$x^2y^3+x(1+y^2)y'=0\qquad\mu (x,y)=\dfrac{1}{xy^3}$$
$$\mbox{Define }M(x,y)=x^2y^3,\quad N(x,y)=x(1+y^2)$$
$$M_y=\dfrac{\partial}{\partial y}[x^2y^3]=3x^2y^2$$
$$N_x=\dfrac{\partial}{\partial x}[x(1+y^2)]=1+y^2$$
$$\mbox{Since }M_y\neq N_x, \mbox{this implies the given equation is not exact. Let's show the given equation multiplied by the integrating factor } \mu (x,y)=\dfrac{1}{xy^3} \mbox{is exact}.$$
$$\dfrac{1}{xy^3}x^2y^3+\dfrac{1}{xy^3}x(1+y^2)y'=x+(y^{-3}+y^{-1})y'=0$$
$$\mbox{Define }M'(x,y)=x,N'(x,y)=y^{-3}+y^{-1}$$
$$M'_y=\dfrac{\partial}{\partial y}{(x)}=0$$
$$M'_x=\dfrac{\partial}{\partial x}[y^{-3}+y^{-1}]=0$$
$$\mbox{Since }M'_y=M'_x, \mbox{this implies } x+(y^{-3}+y^{-1})y'=0 \mbox{ is exact}. \mbox{Thus, there exists a function } \varphi (x,y)$$
$$\mbox{such that}\quad \varphi_x=M, \varphi_y=N$$
\begin{equation*}
\begin{aligned}
\varphi &=\int M\ dx \\
&=\int x\ dx \\
&=\dfrac{1}{2}x^2+h(y)
\end{aligned}
\end{equation*}
$$\varphi_y=h'(y)=y^{-3}+y^{-1}$$
$$h(y)=-\dfrac{1}{2}y^{-2}+ln|y|+C$$
$$\mbox{Therefore},\dfrac{1}{2}x^2-\dfrac{1}{2}y^{-2}+ln|y|=C \mbox{ is the general solution for the given equation}.$$