Author Topic: Q2-T0301  (Read 1563 times)

Victor Ivrii

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Q2-T0301
« on: February 02, 2018, 02:11:11 PM »
Using integrating factor method, solve the given initial value problem and determine at least approximately where the solution is valid.
$$
(2x - y) + (2y - x)y' = 0,\qquad y(1) = 3.
$$

David Chan

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Re: Q2-T0301
« Reply #1 on: February 02, 2018, 02:28:17 PM »
   Let $$M(x, y) = 2x - y \qquad \text{ and } \qquad N(x, y) = 2y - x$$
   Then, $$\frac{\partial}{\partial y}M(x, y) = -1 \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = -1$$
   (Also note that $M$, $N$, $M_y$, $N_x$ are all continuous)  Since $M_y = N_x$, the equation is exact.  Therefore, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= 2x - y = M \tag{1} \\\psi_y(x, y) &= 2y - x = N \tag{2}\end{align*}
   Integrating (1) with respect to $x$, we get $$\psi(x, y) = x^2 - xy + h(y)$$ for some function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = -x + h'(y) = 2y - x$$
   Therefore, $$h'(y) = 2y \implies h(y) = y^2$$
   and we have $$\psi(x, y) = x^2 - xy + y^2$$
   By our choice of $\psi$, we know that $$\frac{\partial \psi}{\partial x} = M \qquad \text{ and } \qquad \frac{\partial \psi}{\partial y} = N$$ so we can rewrite our original differential equation as $$\frac{\partial \psi}{\partial x} + \frac{\partial \psi}{\partial y} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}(\psi(x, y)) = 0$$  Therefore, the (implicit) general solution to out differential equation is $$\psi(x, y) = x^2 - xy + y^2 = C$$
   We can solve this explicitly for $y$ to using the quadratic formula to get $$y = \frac{x \pm \sqrt{-3x^2 + 4C}}{2}$$
   Given the initial condition $y(1) = 3$, we have $$(1)^2 + (1)(3) - (3)^2 = C \implies C = 7$$
   Thus, we have our solution $$y = \frac{x + \sqrt{28 - 3x^2}}{2}, \qquad \text{ valid for $\vert x \vert < \frac{\sqrt{28}}{3}$}$$
« Last Edit: February 02, 2018, 02:58:47 PM by David Chan »