Author Topic: Q2-T0501  (Read 1384 times)

Victor Ivrii

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Q2-T0501
« on: February 02, 2018, 02:12:28 PM »
Show that the given equation is not exact but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$$
(x + 2) \sin(y) + x \cos(y)y' = 0,\qquad  \mu (x, y) = xe^x.
$$

David Chan

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Re: Q2-T0501
« Reply #1 on: February 02, 2018, 02:29:51 PM »
   Let $$M(x, y) = (x + 2)\sin(y) \qquad \text{ and } \qquad  N(x, y) = x\cos(y)$$
   Then, $$\frac{\partial}{\partial y}M(x, y) = (x + 2)\cos(y) \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = \cos(y)$$
   Note that $M_y \neq N_x$, so the equation is not exact.  However, multiplying through by $\mu(x, y) = xe^x$, we get a new equation $$(x + 2)xe^x\sin(y) + x^2e^x\cos(y)y' = 0$$
   We can see that this equation is exact, since $$\frac{\partial}{\partial y} (x + 2)xe^x\sin(y)= (x + 2)xe^x\cos(y) = \frac{\partial}{\partial x} x^2e^x\cos(y)$$
   Therefore, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= (x + 2)xe^x\sin(y) \tag{1} \\\psi_y(x, y) &= x^2e^x\cos(y) \tag{2}\end{align*}
   Integrating (1) with respect to $x$, we get $$\psi(x, y) = \int (x + 2)xe^x\sin(y)\,\mathrm{d}x = x^2e^x\sin(y) + h(y)$$
   for some function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = x^2e^x\cos(y) + h'(y) = x^2e^x\cos(y)$$
   Therefore, $$h'(y) = 0 \implies h \text{ is constant }$$  Taking $h(y) = 0$, we get $$\psi(x, y) = x^2e^x\sin(y)$$
   Thus, the solutions of the differential equation are given implicitly by $$x^2e^x\sin(y) = C$$
« Last Edit: February 02, 2018, 04:14:02 PM by David Chan »