Math Question and Answers


Question and Answers Forum

All Questions Topic List
Differential Equation Questions
Previous in All Question Next in All Question
Previous in Differential Equation Next in Differential Equation
Question Number 67991 by ramirez105 last updated on 03/Sep/19

Commented by mathmax by abdo last updated on 03/Sep/19

$x y d x + 2 (x^{2} + 2 y^{2}) d y = 0 \Rightarrow 2 (x^{2} + 2 y^{2}) d y = - x y d x \Rightarrow$ $2 \frac{d y}{y} = \frac{- x d x}{x^{2} + 2 y^{2}} \Rightarrow \int \frac{2 d y}{y} = - \int \frac{x d x}{x^{2} + 2 y^{2}} + c \Rightarrow$ $2 l n ∣ y ∣ = - \frac{1}{2} l n (x^{2} + 2 y^{2}) + c \Rightarrow y^{2} = \frac{k}{\sqrt{x^{2} + 2 y^{2}}} \Rightarrow$ $y^{4} = \frac{k^{2}}{x^{2} + 2 y^{2}} \Rightarrow x^{2} y^{4} + 2 y^{6} - k^{2} = 0 \Rightarrow 2 y^{6} + x^{2} y^{4} - k^{2} = 0$ $l e t y^{2} = t \Rightarrow 2 t^{3} + x^{2} t^{2} - k^{2} = 0 . . . b e c o n t i n u e d . .$
	Answered by Tanmay chaudhury last updated on 03/Sep/19
	$(dy/dx)=−(((xy)/(2x^2 +4y^2 ))) (dy/dx)=−(((y/x)/(2+(y^2 /x^2 )))) now put (y/x)=v →y=vx (dy/dx)=v+x(dv/dx) v+x(dv/dx)=−((v/(2+v^2 ))) x×(dv/dx)=((−v)/(2+v^2 ))−v x×(dv/dx)=((−v−2v−v^3 )/(2+v^2 )) ((2+v^2 )/(3v+v^3 ))dv=((−dx)/x) ((2+v^2 )/(v(3+v^2 )))=((−dx)/x) ((3+v^2 −1)/(v(3+v^2 )))dv=((−dx)/x) (dv/v)−(dv/(v(3+v^2 )))=((−dx)/x) (dv/v)−(1/3)×((3+v^2 −v^2 )/(v(3+v^2 )))dv=((−dx)/x) (dv/v)−(1/3)×(dv/v)+(1/3)×(v/(3+v^2 ))dv=((−dx)/x) (2/3)×(dv/v)+(1/6)×((d(3+v^2 ))/(3+v^2 ))=((−dx)/x) intrgating both side (2/3)∫(dv/v)+(1/6)∫((d(3+v^2 ))/(3+v^2 ))=−∫(dx/x) (2/3)lnv+(1/6)ln(3+v^2 )=−lnx +lnC ln{v^(2/3) .(3+v^2 )^(1/6) }.+lnx=lnC x.v^(2/3) .(3+v^2 )^(1/6) =C x.((y/(x )))^(2/3) .(3+(y^2 /x^2 ))^(1/6) =C x^(1−(2/3)−(2/6)) ×y^(2/3) ×(3x^2 +y^2 )^(1/6) =C y^(2/3) ×(3x^2 +y^2 )^(1/6) =C putting x=0 y=1 we get C=1 y^(2/3) ×(3x^2 +y^2 )^(1/6) =1 y^4 (3x^2 +y^2 )=1 answer$
	$\frac{d y}{d x} = - (\frac{x y}{2 x^{2} + 4 y^{2}})$ $\frac{d y}{d x} = - (\frac{\frac{y}{x}}{2 + \frac{y^{2}}{x^{2}}}) n o w p u t \frac{y}{x} = v \to y = v x$ $\frac{d y}{d x} = v + x \frac{d v}{d x}$ $v + x \frac{d v}{d x} = - (\frac{v}{2 + v^{2}})$ $x \times \frac{d v}{d x} = \frac{- v}{2 + v^{2}} - v$ $x \times \frac{d v}{d x} = \frac{- v - 2 v - v^{3}}{2 + v^{2}}$ $\frac{2 + v^{2}}{3 v + v^{3}} d v = \frac{- d x}{x}$ $\frac{2 + v^{2}}{v (3 + v^{2})} = \frac{- d x}{x}$ $\frac{3 + v^{2} - 1}{v (3 + v^{2})} d v = \frac{- d x}{x}$ $\frac{d v}{v} - \frac{d v}{v (3 + v^{2})} = \frac{- d x}{x}$ $\frac{d v}{v} - \frac{1}{3} \times \frac{3 + v^{2} - v^{2}}{v (3 + v^{2})} d v = \frac{- d x}{x}$ $\frac{d v}{v} - \frac{1}{3} \times \frac{d v}{v} + \frac{1}{3} \times \frac{v}{3 + v^{2}} d v = \frac{- d x}{x}$ $\frac{2}{3} \times \frac{d v}{v} + \frac{1}{6} \times \frac{d (3 + v^{2})}{3 + v^{2}} = \frac{- d x}{x}$ $i n t r g a t i n g b o t h s i d e$ $\frac{2}{3} \int \frac{d v}{v} + \frac{1}{6} \int \frac{d (3 + v^{2})}{3 + v^{2}} = - \int \frac{d x}{x}$ $\frac{2}{3} l n v + \frac{1}{6} l n (3 + v^{2}) = - l n x + l n C$ $l n {v^{\frac{2}{3}} . {(3 + v^{2})}^{\frac{1}{6}}} . + l n x = l n C$ $x . v^{\frac{2}{3}} . {(3 + v^{2})}^{\frac{1}{6}} = C$ $x . {(\frac{y}{x})}^{\frac{2}{3}} . {(3 + \frac{y^{2}}{x^{2}})}^{\frac{1}{6}} = C$ $x^{1 - \frac{2}{3} - \frac{2}{6}} \times y^{\frac{2}{3}} \times {(3 x^{2} + y^{2})}^{\frac{1}{6}} = C$ $y^{\frac{2}{3}} \times {(3 x^{2} + y^{2})}^{\frac{1}{6}} = C$ $p u t t i n g x = 0 y = 1 w e g e t C = 1$ $y^{\frac{2}{3}} \times {(3 x^{2} + y^{2})}^{\frac{1}{6}} = 1$ $y^{4} (3 x^{2} + y^{2}) = 1 a n s w e r$
	Commented by Prithwish sen last updated on 03/Sep/19

	$Where have you been so long sir ?$
	Commented by peter frank last updated on 03/Sep/19

	$t h a n k y o u$
	Commented by Tanmay chaudhury last updated on 03/Sep/19

	$s i r i a m a l s o i n v o l v e d i n a n o t h e r f o r u m g o i i t . .$
	Commented by mr W last updated on 03/Sep/19

	$p l e a s e c h e c k s i r :$ $\frac{d y}{d x} = - (\frac{\frac{y}{x}}{2 + 4 \frac{y^{2}}{x^{2}}}) n o w p u t \frac{y}{x} = v \to y = v x$ $. . .$ $v + x \frac{d v}{d x} = - (\frac{v}{2 + 4 v^{2}})$ $. . .$
	Commented by Tanmay chaudhury last updated on 03/Sep/19

	$y e s s i r . . y o u a r e r i g h t . . f o r g o t t o p u t 4 . . . t h a n k y o u s i r . . .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum