Question

solve (x²+y²)dx - (xy -x²) dy =0​

Answer 1

$\large\underline{\sf{Solution-}}$

The given Differential equation is

$\rm :\longmapsto\:( {x}^{2} + {y}^{2})dx - (xy - {x}^{2})dy = 0$

can be rewritten as .

$\rm :\longmapsto\:( {x}^{2} + {y}^{2})dx = (xy - {x}^{2})dy$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ {x}^{2} + {y}^{2} }{ xy - {x}^{2} }$

Since, degree of numerator and denominator both are 2.

So, its a homogeneous differential equation of order 0.

So, to evaluate this Differential equation,

$\purple{\bf :\longmapsto\:Put \: y \: = \: vx} - - - (1)$

So, given differential equation reduces to

$\rm :\longmapsto\:\dfrac{d}{dx}vx= \dfrac{ {x}^{2} + {(vx)}^{2} }{x(vx) - {x}^{2} }$

$\rm :\longmapsto\:v\dfrac{d}{dx}x + x\dfrac{d}{dx}v =\dfrac{ {x}^{2} + {v}^{2} {x}^{2} }{ {vx}^{2} - {x}^{2} }$

$\rm :\longmapsto\:v + x\dfrac{dv}{dx} = \dfrac{ {x}^{2}(1 + {v}^{2} )}{ {x}^{2}(v - 1)}$

$\rm :\longmapsto\:v + x\dfrac{dv}{dx} = \dfrac{ 1 + {v}^{2} }{v - 1}$

$\rm :\longmapsto\: x\dfrac{dv}{dx} = \dfrac{ 1 + {v}^{2} }{v - 1} - v$

$\rm :\longmapsto\: x\dfrac{dv}{dx} = \dfrac{ 1 + {v}^{2} - {v}^{2} + v }{v - 1}$

$\rm :\longmapsto\: x\dfrac{dv}{dx} = \dfrac{ 1 + v }{v - 1}$

On separate the variables, we get

$\rm :\longmapsto\: \dfrac{dx}{x} = \dfrac{ v - 1}{v + 1}dv$

On integrating, both sides,

$\rm :\longmapsto\:\displaystyle\int\tt \dfrac{dx}{x} = \displaystyle\int\tt \dfrac{ v - 1}{v + 1}dv$

We know,

$\boxed{ \sf{ \: \displaystyle\int\tt \frac{dx}{x} = log(x) + c}}$

So, using this

$\rm :\longmapsto\:logx = \displaystyle\int\tt \dfrac{ v + 1 - 1 - 1}{v + 1}dv$

$\rm :\longmapsto\:logx = \displaystyle\int\tt \dfrac{ v + 1 - 2}{v + 1}dv$

$\rm :\longmapsto\:logx =\displaystyle\int\tt 1dv - 2 \displaystyle\int\tt \dfrac{1}{v + 1}dv$

$\rm :\longmapsto\:logx = v - 2log(v + 1) + c$

On substituting the value of v, from equation (1), we get

$\rm :\longmapsto\:logx = \dfrac{y}{x} - 2log \bigg(\dfrac{y}{x} + 1 \bigg) + c$

$\rm :\longmapsto\:logx = \dfrac{y}{x} - 2log \bigg(\dfrac{y + x}{x}\bigg) + c$

$\rm :\longmapsto\:logx = \dfrac{y}{x} - 2log (y + x) + 2logx + c$

$\rm :\longmapsto\: - logx - \dfrac{y}{x} + 2log (y + x) = c$

Answer 2

Answer:

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Step-by-step explanation:

Correct option is

Bx 2

(x 2 −2y 2 )=k

(x 2 −y 2 )dx−xydy=0

⇒ dxdy

= xyx 2 −y 2

which is homogeneous differential eqn.

Put y=vx DXdy

=v+x dxdv

So, the eqn becomes

⇒v+x dxd = v1−v 2

⇒x dxdv= v1−2v 2

⇒ 1−2v 2v

dv= xdx ⇒∫ 1−2v 2v

dv=∫ xdx Put 1−2v 2 =t⇒−4vdv=dt⇒− 41

∫ tdt =logx+log ⇒−− 41

log∣1−2v 2 ∣=logCx⇒ 41

log∣ x 2 −2y 2x 2

∣=logCx⇒∣ x 2 −2y 2x 2

∣=cx 4⇒x 2

(x 2 −2y 2 )= c1

⇒x 2 - 2(x 2 −2y 2 )=k

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