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MAT244--2019F => MAT244--Test & Quizzes => Quiz-2 => Topic started by: Gavrilo Milanov Dzombeta on October 04, 2019, 02:00:43 PM

Title: Q2: TUT0102
Post by: Gavrilo Milanov Dzombeta on October 04, 2019, 02:00:43 PM
$$\text{Find an integrating factor and solve the following equation: }$$
$$e^{x} + \left(e^{x} \cot(y) + 2y \csc(y)\right)y^\prime = 0 \tag{1}$$

$$\text{Let } M = e^{x} \text{ and } N = e^{x} \cot(y) + 2y \csc(y)$$
$$\text{Then } M_y = 0 \text{ and } N_x = e^{x} \cot(y)$$
$$M_y \neq N_x \implies \text{The equation is not exact.}$$
$$\dfrac{N_x - M_y}{M} = \dfrac{e^{x}\cot(y)}{e^{x}} = \cot(y)$$
$$\mu = e^{\int{\cot(y)dy}} = e^{\int \frac{\cos(y)}{\sin(y)}dy}$$
$$\text{The integrating factor is: } \mu = \sin(y)$$
$$\text{Multiply equation 1 by \mu.}$$
$${e^{x}}\sin(y) + \left({e^{x}}\cos(y) + 2y\right)y^\prime = 0 \tag{2}$$
$$\text{ Let } \tilde{M} = e^{x}\sin(y) \text{ and } \tilde{N} = {e^{x}}\cos(y) + 2y$$
$$\tilde{M}_y = {e^{x}}\cos(y) \text{ and } \tilde{N}_x = {e^{x}}\cos(y)$$
$$\tilde{M}_y = \tilde{N}_x \implies \text{ Equation 2 is exact.}$$
$$\psi_x = M \tag{3}$$
$$\text{Integrating equation 3 with respect to x},$$
$$\int{\psi_x dx} = \int{e^{x} \sin(y) dx}$$
$$\therefore \psi = e^{x} \sin(y) + h(y) \tag{4}$$
$$\psi_y = e^{x} \cos(y) + h^\prime(y)$$
$$\psi_y = N$$
$$\therefore e^{x} \cos(y) + h^\prime(y) = e^{x}\cos(y) + 2y$$
$$\int{h^\prime(y) dy} = \int{2y dy} \implies h(y) = y^{2}$$
$$\therefore e^{x}\sin(y) + y^{2} = c$$