(y^2 +2) dx = (xy+2y+y^3 ) dy


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Question Number 103931 by bemath last updated on 18/Jul/20

$(y^{2} + 2) d x = (x y + 2 y + y^{3}) d y$
	Answered by bobhans last updated on 18/Jul/20

	$\frac{d y}{d x} = \frac{y^{2} + 2}{y (x + 2 + y^{2})}$ $s e t : q = x + 2 + y^{2} \Rightarrow \frac{d q}{d x} = 1 + 2 y \frac{d y}{d x}$ $2 y \frac{d y}{d x} = \frac{d q}{d x} - 1 \Rightarrow \frac{d y}{d x} = \frac{1}{2 y} \frac{d q}{d x} - \frac{1}{2 y}$ $\Leftrightarrow \frac{1}{2 y} \frac{d q}{d x} - \frac{1}{2 y} = \frac{q - x}{q y}$ $\frac{1}{2} (\frac{d q}{d x} - 1) = 1 - \frac{x}{q}$ $\frac{d q}{d x} - 1 = 2 - \frac{2 x}{q} \Rightarrow \frac{d q}{d x} = 3 - \frac{2 x}{q}$ $\frac{d q}{d x} + \frac{2 x}{q} = 3$
	Answered by bramlex last updated on 18/Jul/20
	$(∂M/∂y) = 2y ; (∂N/∂x) = −y ⇒non exact (∂N/∂x)−(∂M/∂y) = −3y is a function only in y . integrating factor u(y)=e^(∫(((−3y)/(y^2 +2))) dy) u(y)= e^(−(3/2)∫ ((d(y^2 +2))/(y^2 +2))) = (1/((y^2 +2)^(3/2) )) ⇒(((y^2 +2)/(√((y^2 +2)^3 )))) dx − (((xy+2y+y^3 )/(√((y^2 +2)^3 )))) dy=0 (1/(√(y^2 +2))) dx − ((x+2y+y^3 )/(√(y^2 +2))) dy=0 { (((∂f/∂x) = M(x,y)=(1/(√(y^2 +2))))),(((∂f/∂y)=N(x,y)=((x+2y+y^3 )/(√((y^2 +2)^3 ))))) :} F(x,y)=∫M(x,y)dx+g(y) g′(y) =((−y)/(√(y^2 +2))) g(y) = −(1/2)∫ ((d(y^2 +2))/((y^2 +2)^(1/2) )) g(y) = (√(y^2 +2)) F(x,y)= ∫ (dx/(√(y^2 +2))) + (√(y^2 +2)) F(x,y) = (x/(√(y^2 +2))) +(√(y^2 +2)) F(x,y) = ((x+y^2 +2)/(√(y^2 +2))) .★$
	$\frac{\partial M}{\partial y} = 2 y; \frac{\partial N}{\partial x} = - y \Rightarrow n o n e x a c t$ $\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = - 3 y i s a f u n c t i o n o n l y i n y .$ $i n t e g r a t i n g f a c t o r u (y) = e^{\int (\frac{- 3 y}{y^{2} + 2}) d y}$ $u (y) = e^{- \frac{3}{2} \int \frac{d (y^{2} + 2)}{y^{2} + 2}} = \frac{1}{{(y^{2} + 2)}^{3 / 2}}$ $\Rightarrow (\frac{y^{2} + 2}{\sqrt{{(y^{2} + 2)}^{3}}}) d x - (\frac{x y + 2 y + y^{3}}{\sqrt{{(y^{2} + 2)}^{3}}}) d y = 0$ $\frac{1}{\sqrt{y^{2} + 2}} d x - \frac{x + 2 y + y^{3}}{\sqrt{y^{2} + 2}} d y = 0$ ${\begin{cases} \frac{\partial f}{\partial x} = M (x, y) = \frac{1}{\sqrt{y^{2} + 2}} \\ \frac{\partial f}{\partial y} = N (x, y) = \frac{x + 2 y + y^{3}}{\sqrt{{(y^{2} + 2)}^{3}}} \end{cases}$ $F (x, y) = \int M (x, y) d x + g (y)$ $g^{'} (y) = \frac{- y}{\sqrt{y^{2} + 2}}$ $g (y) = - \frac{1}{2} \int \frac{d (y^{2} + 2)}{{(y^{2} + 2)}^{1 / 2}}$ $g (y) = \sqrt{y^{2} + 2}$ $F (x, y) = \int \frac{d x}{\sqrt{y^{2} + 2}} + \sqrt{y^{2} + 2}$ $F (x, y) = \frac{x}{\sqrt{y^{2} + 2}} + \sqrt{y^{2} + 2}$ $F (x, y) = \frac{x + y^{2} + 2}{\sqrt{y^{2} + 2}} . ★$
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