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Question Number 121461 by bramlexs22 last updated on 08/Nov/20

	Answered by Olaf last updated on 08/Nov/20
	$(3y^2 +10xy^2 )dx+(6xy−2+10x^2 y)dy = 0 df = (∂f/∂x)(x,y)dx+(∂f/∂y)(x,y)dy = 0 with : (∂f/∂x)(x,y) = 3y^2 +10xy ⇒ f(x,y) = 3y^2 x+5x^2 y^2 +A(y) and : (∂f/∂y)(x,y) = 6xy−2+10x^2 y ⇒ f(x,y) = 3xy^2 +5x^2 y^2 +B(x) df = 0 ⇒ f(x,y) = C_1 { ((3xy^2 +5x^2 y^2 +A(y) = C_1 (1))),((3xy^2 +5x^2 y^2 +B(x) = C_1 (2))) :} (1)−(2) : ∀x,y, A(y)−B(x) = 0 ⇒ A(y) = B(x) = C_2 Finally f(x,y) = 3xy^2 +5x^2 y^2 +C_2 = C_1 ⇒ y^2 = (C_3 /(x(3+5x)))$
	$(3 y^{2} + 10 {x y}^{2}) d x + (6 x y - 2 + 10 x^{2} y) d y = 0$ $d f = \frac{\partial f}{\partial x} (x, y) d x + \frac{\partial f}{\partial y} (x, y) d y = 0$ $with :$ $\frac{\partial f}{\partial x} (x, y) = 3 y^{2} + 10 x y$ $\Rightarrow f (x, y) = 3 y^{2} x + 5 x^{2} y^{2} + A (y)$ $and :$ $\frac{\partial f}{\partial y} (x, y) = 6 x y - 2 + 10 x^{2} y$ $\Rightarrow f (x, y) = 3 {x y}^{2} + 5 x^{2} y^{2} + B (x)$ $d f = 0 \Rightarrow f (x, y) = C_{1}$ ${\begin{cases} 3 {x y}^{2} + 5 x^{2} y^{2} + A (y) = C_{1} (1) \\ 3 {x y}^{2} + 5 x^{2} y^{2} + B (x) = C_{1} (2) \end{cases}$ $(1) - (2) : \forall x, y, A (y) - B (x) = 0$ $\Rightarrow A (y) = B (x) = C_{2}$ $Finally f (x, y) = 3 {x y}^{2} + 5 x^{2} y^{2} + C_{2} = C_{1}$ $\Rightarrow y^{2} = \frac{C_{3}}{x (3 + 5 x)}$
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