Question

Solve the Differential Equation

(x^2 - y^2)dx + 2xydy = 0

please give correct answer ​

Answer 1

Answer ----> ( x² + y² ) = cx

Given----> ( x² - y² ) dx + 2xy dy = 0

To find-----> Solve given differential equation

Solution----> ATQ,

( x² - y² ) dx + 2xy dy = 0

=> 2xy dy = - ( x² - y² ) dx

=> dy / dx = - ( x² - y² ) / 2xy ...................( 1 )

This is a homogenous given differential equation.

So , putting , y = vx

Now differentiating with respect to x .

dy / dx = d / dx ( vx )

Applying product rule of differentiation , we get,

=> dy/dx = v d/dx ( x ) + x dv/dx

=> dy/dx = v ( 1 ) + x dv/dx

=> dy/dx = v + x dv/dx

Now by ( 1 ) , we get,

v + x dv/dx = - ( x² - v²x² ) / 2x ( vx )

= - x² ( 1 - v² ) / 2x² v

x² Cancel out from numerator and denominator in RHS, we get.

v + x dv/dx = - ( 1 - v² ) / 2v

=> x dv/dx = { ( v² - 1 ) / 2v } - v

= ( v² - 1 - 2v² ) / 2v

= ( -1 - v² ) / 2v

=> x dv/dx = - ( 1 + v² ) / 2v

=> 2v dv / ( 1 + v² ) = - dx / x

Integrating both sides we get ,

=> ∫ 2v dv / ( 1 + v² ) = - ∫ dx / x + log c ...........( 2 )

Let, 1 + v² = t

Differentiating with respect to x , we get,

=> (0 + 2v ) dv = dt

=> 2v dv = dt

So , by ( 2 ) , we get,

=> ∫ dt / t = - logx + logc

=> log t = log ( c / x )

=> t = c / x

=> t x = c

Putting t = 1 + v² , in it , we get,

=> ( 1 + v² ) x = c

Now putting v = y / x , in it , we get,

=> { 1 + ( y / x )² } x = C

=> { 1 + ( y² / x² ) } x = c

=> { ( x² + y² ) ) x² } x = c

=> ( x² + y² ) / x = c

=> ( x² + y² ) = cx

Answer 2

Given, (x2+y2)dx−2xydy=0

⇒(x2+y2)dx=2xydy

⇒dxdy=2xyx2+y2    .... (i)

Let y=vx

Thus, dxdy=v+xdxdv

Thus, v+xdxdv=2x(vx)x2+(vx)2

⇒v+xdxdv=2v1+v2

⇒xdxdv=2v1+v2−v

⇒xdxdv=2v1+v2−2v2