### Author Topic: Q2-T0501  (Read 2037 times)

#### Victor Ivrii ##### 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

• Jr. Member
•  • Posts: 13
• Karma: 8 ##### 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 »