Resolve 1) x(dy/dx)−y=y^3 2) (x−y)ydx−x^2 dy=0 3) (2x−y)dx+(4x−2y+3)dy=0


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Question Number 168613 by LEKOUMA last updated on 14/Apr/22

$R e s o l v e$ $1) x \frac{d y}{d x} - y = y^{3}$ $2) (x - y) y d x - x^{2} d y = 0$ $3) (2 x - y) d x + (4 x - 2 y + 3) d y = 0$
Commented by Mastermind last updated on 14/Apr/22

$S o l u t i o n 1$ $x \frac{d y}{d x} - y = y^{3}$ $\Rightarrow \frac{1}{y^{3}} \frac{d y}{d x} - \frac{1}{{x y}^{2}} = \frac{1}{x} . . . . . . . . . (1)$ $p u t z = - \frac{1}{y^{2}}, s o t h a t \frac{d z}{d x} = 2 \frac{1}{y^{3}} \frac{d y}{d x}$ $\Rightarrow \frac{1}{2} \frac{d z}{d x} = \frac{1}{y^{3}} \frac{d y}{d x}$ $f r o m (1), w e h a v e$ $\frac{1}{2} \frac{d z}{d x} - \frac{z}{x} = \frac{1}{x}$ $m u l t i p l y a l l t h r o u g h b y 2, t h e n$ $\frac{d z}{d x} - \frac{2 z}{x} = \frac{2}{x} . . . . . . . . (2)$ $N o w, I F = e^{\int - \frac{2}{x}} \Rightarrow e^{l n (x^{- 2})}$ $\Rightarrow x^{- 2} = \frac{1}{x^{2}}$ $m u l t i p l y b o t h s i d e s b y I F$ $w e h a v e, \frac{1}{x^{2}} (\frac{d z}{d x} - \frac{2 z}{x}) = \frac{2}{x^{3}}$ $N o w, \frac{d}{d x} (z \cdot \frac{1}{x^{2}}) = \frac{2}{x^{3}}$ $d (z \cdot \frac{1}{x^{2}}) = \frac{2}{x^{3}} d x$ $O n i n t e g r a t i n g, w e g e t$ $\frac{z}{x^{2}} = - \frac{1}{x^{2}} + c$ $\Rightarrow z = - 1 + {c x}^{2}$ $p u t t i n g z = - \frac{1}{y^{2}}$ $W e h a v e,$ $- \frac{1}{y^{2}} = - 1 + {c x}^{2}$ $\Rightarrow \frac{1}{y^{2}} = 1 - {c x}^{2}$ $R e s o l v e d$ $M a s t e r m i n d$
	Answered by mr W last updated on 15/Apr/22

	$(1)$ $\frac{d y}{y (1 + y^{2})} = \frac{d x}{x}$ $(\frac{1}{y} - \frac{y}{1 + y^{2}}) d y = \frac{d x}{x}$ $\ln y - \frac{1}{2} \ln (1 + y^{2}) = \ln x + C$ $\Rightarrow \frac{y^{2}}{1 + y^{2}} = {C x}^{2}$ $o r$ $\Rightarrow (1 - {C x}^{2}) (1 + y^{2}) = 1$ $o r$ $\Rightarrow \frac{1}{1 + y^{2}} = 1 - {C x}^{2}$
	Answered by mr W last updated on 15/Apr/22

	$(2)$ $\frac{d y}{d x} = \frac{(x - y) y}{x^{2}}$ $y = x t$ $\frac{d y}{d x} = t + x \frac{d t}{d x}$ $t + x \frac{d t}{d x} = t (1 - t)$ $x \frac{d t}{d x} = - t^{2}$ $\frac{d t}{t^{2}} = - \frac{d x}{x}$ $\int \frac{d t}{t^{2}} = - \int \frac{d x}{x}$ $- \frac{1}{t} = - \ln x - C$ $\frac{x}{y} = \ln x + C$ $\Rightarrow y = \frac{x}{\ln x + C}$
	Answered by mr W last updated on 15/Apr/22

	$(3)$ $\frac{d y}{d x} = \frac{- 2 x + y}{4 x - 2 y + 3}$ $l e t u = - 2 x + y$ $\frac{d u}{d x} = - 2 + \frac{d y}{d x}$ $2 + \frac{d u}{d x} = \frac{u}{- 2 u + 3}$ $\frac{d u}{d x} = \frac{5 u - 6}{- 2 u + 3}$ $\frac{(2 u - 3) d u}{5 u - 6} = - d x$ $\frac{(10 u - 12 - 3) d u}{5 u - 6} = - 5 d x$ $(2 - \frac{3}{5 u - 6}) d u = - 5 d x$ $2 u - \frac{3}{5} \ln (5 u - 6) = - 5 x + C$ $- 4 x + 2 y - \frac{3}{5} \ln (- 10 x + 5 y - 6) = - 5 x + C$ $\Rightarrow x + 2 y - \frac{3}{5} \ln (- 10 x + 5 y - 6) = C$
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