### Author Topic: Q2-T0401  (Read 1367 times)

#### Victor Ivrii

• Elder Member
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##### Q2-T0401
« on: February 02, 2018, 02:11:57 PM »
Find an integrating factor and solve the given equation.
$$1 + \bigl(\frac{x}{y} - \sin(y)\bigr)y' = 0.$$

#### David Chan

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##### Re: Q2-T0401
« Reply #1 on: February 02, 2018, 02:28:58 PM »
Let $$M(x, y) = 1 \qquad \text{ and } \qquad N(x, y) = \left(\frac{x}{y} - \sin(y)\right)$$
Then, $$\frac{\partial}{\partial y}M(x, y) = 0 \qquad \text{ and } \qquad \frac{\partial}{\partial x}N(x, y) = \frac{1}{y}$$
We can see that this equation is not exact, however, note that $$\frac{N_x - M_y}{M} = \frac{1}{y}$$ is a function of $y$ only.  Thus, there is an integrating factor $\mu(y)$ that satisfies the differential equation $$\frac{\mathrm{d}\mu}{\mathrm{d}y} = \left(\frac{N_x - M_y}{M}\right)\mu = \frac{\mu}{y} \qquad \implies \qquad \mu = y$$
Multiplying our original equation by $\mu(y)$, we have $$y + (x - y\sin(y))y' = 0$$
We can see that this equation is exact, since $$\frac{\partial}{\partial y}(y) = 1 = \frac{\partial}{\partial x}(x - y\sin(y))$$
Thus, there exists a function $\psi(x, y)$ such that \begin{align*}\psi_x(x, y) &= y \tag{1} \\\psi_y(x, y) &= x - y\sin(y) \tag{2}\end{align*}
Integrating (1) with respect to $x$, we get $$\psi(x, y) = xy + h(y)$$ for some arbitrary function $h$ of $y$.  Next, differentiating with respect to $y$, and equating with (2), we get $$\psi_y(x, y) = x + h'(y)$$
Therefore, $$h'(y) = -y\sin(y) \qquad \implies \qquad h(y) = -\int y \sin(y)\,\mathrm{d}y = y\cos(y) - \sin(y)$$
and we have $$\psi(x, y) = xy + y\cos(y) - \sin(y)$$
Thus, the solutions of the differential equation are given implicitly by $$xy + y\cos(y) - \sin(y) = C$$