Solve:
(𝑥 − 2𝑠𝑖𝑛𝑦 + 3)𝑑𝑥 + (2𝑥 − 4𝑠𝑖𝑛𝑦 − 3)𝑐𝑜𝑠𝑦𝑑𝑦 = 0
(x−2siny+3)dx+(2x−4siny−3)cosydy=0u=sinydu=cosydy⇒dy=du/cosy\begin{gathered} (x-2 \sin y+3) d x+(2 x-4 \sin y-3) \cos y d y=0 \\ u=\sin y \\ d u=\cos y d y \Rightarrow d y=d u / \cos y \end{gathered}(x−2siny+3)dx+(2x−4siny−3)cosydy=0u=sinydu=cosydy⇒dy=du/cosy
Substituting in differential equation.
(x−2u+3)dx+(2x−4u−3)cosyducosy=0 Integrating ⇒∫(x−2u+3)dx+∫(2x−4u−3)du=C⇒x22−2ux+3x+2x22−4ux−3u=C⇒3x2−64x−3u+3x=C2⇒3x23−6xsiny−3siny+3x=C\begin{aligned} &(x-2 u+3) d x+(2 x-4 u-3) \cos y \frac{d u}{\cos y}=0 \\ & \text { Integrating } \\ \Rightarrow & \int(x-2 u+3) d x+\int(2 x-4 u-3) d u=C \\ \Rightarrow & \frac{x^{2}}{2}-2u x+3 x+\frac{2 x^{2}}{2}-4u x-3 u=C \\ \Rightarrow & \frac{3 x^{2}-64 x-3 u+3 x=C}{2}\\\Rightarrow & \frac{3 x^{2}}{3}-6 x \sin y-3 \sin y+3 x=C \end{aligned}⇒⇒⇒⇒(x−2u+3)dx+(2x−4u−3)cosycosydu=0 Integrating ∫(x−2u+3)dx+∫(2x−4u−3)du=C2x2−2ux+3x+22x2−4ux−3u=C23x2−64x−3u+3x=C33x2−6xsiny−3siny+3x=C
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