MAT244-2018S > Quiz-2

Q2-T0401

(1/1)

**Victor Ivrii**:

Find an integrating factor and solve the given equation.

$$

1 + \bigl(\frac{x}{y} - \sin(y)\bigr)y' = 0.

$$

**David Chan**:

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 $$xy + y\cos(y) - \sin(y) = C$$

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