Author Topic: Q2-T0101  (Read 1695 times)

Victor Ivrii

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Q2-T0101
« on: February 02, 2018, 02:09:49 PM »
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$
x^2y^3 + x(1 + y^2)y' = 0,\qquad  \mu(x, y) = 1/xy^3.
$$

David Chan

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Re: Q2-T0101
« Reply #1 on: February 02, 2018, 02:26:19 PM »
Let $$M(x, y) = x^2 y^3 \qquad \text{ and } \qquad  N(x, y) = x(1 + y^2)$$
   Then, $$\frac{\partial}{\partial y}M(x, y) = 3x^2y^2 \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = 1 + y^2$$
   Note that $M_y \neq N_x$, so the equation is not exact.  Note also that $y \equiv 0$ is a solution.  Supposing $x, y \neq 0$, we can multiply through by $\mu(x, y) = \frac{1}{xy^3}$, we get a new equation $$x + \frac{1 + y^2}{y^3}y' = 0$$    
   We can see that this equation is exact, since $$\frac{\partial}{\partial y}(x) = 0 = \frac{\partial}{\partial x}\left(\frac{1 + y^2}{y^3}\right)$$
   Therefore, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= x \tag{1} \\\psi_y(x, y) &= \frac{1 + y^2}{y^2} \tag{2}\end{align*}
   Integrating (1) with respect to $x$, we get $$\psi(x, y) = \frac{1}{2}x^2 + h(y)$$ for some function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = h'(y) = \frac{1 + y^2}{y^3}$$
   Therefore, $$h'(y) = \frac{1 + y^2}{y^3} \implies h(y) = \int \frac{\mathrm{d}y}{y^3} + \int \frac{\mathrm{d}y}{y}= \log{|y|} - \frac{1}{2y^2}$$
   and we have $$\psi(x, y) = \frac{1}{2}x^2 + \log{|y|} - \frac{1}{2y^2}$$
   Thus, the solutions of the differential equation are given implicitly by $$\frac{1}{2}x^2 + \log{|y|} - \frac{1}{2y^2} = C \qquad \text{ or } \qquad y \equiv 0$$
« Last Edit: February 02, 2018, 04:08:05 PM by David Chan »

Victor Ivrii

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Re: Q2-T0101
« Reply #2 on: February 02, 2018, 06:07:57 PM »
David

Your solution and posting is perfect, but you should not poach :D