Author Topic: Q2: TUT0102  (Read 297 times)

Gavrilo Milanov Dzombeta

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Q2: TUT0102
« on: October 04, 2019, 02:00:43 PM »
$$ \text{Find an integrating factor and solve the following equation: }$$
$$ e^{x} + \left(e^{x} \cot(y) + 2y \csc(y)\right)y^\prime = 0 \tag{1} $$
$$ $$
$$ \text{Let } M = e^{x} \text{ and } N = e^{x} \cot(y) + 2y \csc(y) $$
$$ \text{Then } M_y = 0 \text{ and } N_x = e^{x} \cot(y)$$
$$ M_y \neq N_x \implies \text{The equation is not exact.}$$
$$ \dfrac{N_x - M_y}{M} = \dfrac{e^{x}\cot(y)}{e^{x}} = \cot(y)$$
$$ \mu = e^{\int{\cot(y)dy}} = e^{\int \frac{\cos(y)}{\sin(y)}dy} $$
$$ \text{The integrating factor is: } \mu = \sin(y) $$
$$ \text{Multiply equation 1 by $\mu$.}$$
$$ {e^{x}}\sin(y) + \left({e^{x}}\cos(y) + 2y\right)y^\prime = 0 \tag{2}$$
$$ \text{ Let } \tilde{M} = e^{x}\sin(y) \text{ and } \tilde{N} = {e^{x}}\cos(y) + 2y $$
$$ \tilde{M}_y = {e^{x}}\cos(y) \text{ and } \tilde{N}_x = {e^{x}}\cos(y) $$
$$ \tilde{M}_y = \tilde{N}_x \implies \text{ Equation 2 is exact.} $$
$$ \psi_x  = M \tag{3} $$
$$ \text{Integrating equation 3 with respect to $x$}, $$
$$ \int{\psi_x dx} = \int{e^{x} \sin(y) dx} $$
$$ \therefore \psi = e^{x} \sin(y) + h(y) \tag{4} $$
$$ \psi_y = e^{x} \cos(y) + h^\prime(y)$$
$$ \psi_y = N$$
$$ \therefore e^{x} \cos(y) + h^\prime(y) = e^{x}\cos(y) + 2y $$
$$ \int{h^\prime(y) dy} = \int{2y dy} \implies h(y) = y^{2}$$
$$ \therefore e^{x}\sin(y) + y^{2} = c $$