AP EAMCET2003MathematicsDifferential Equations
The general solution of y^2 d x+ (x^2-x y+y^2 ) d y=0 is :
Options
- A⁻¹ ( y x )= y+C
- B2 ⁻¹ ( x y )+ x+C=0
- C(y+ x^2+y^2 )+ y+C=0
- D⁻¹ ( x y )+ y+C=0
Correct answer
A. ⁻¹ ( y x )= y+C
Step-by-step solution
We have, y^2 d x+ (x^2-x y+y^2 ) d y=0 d y d x = -y^2 x^2-x y+y^2 It is a homogeneous linear differential equation Put y=v x d y d x =v+x d v d x v+x d v d x = -v^2 x^2 x^2-v x^2+x^2 v^2 = -v^2 v^2-v+1 x d v d x = -v^2-v^3+v^2-v v^2-v+1 = -v^3-v v^2-v+1 (v^2-v+1 ) -v^3-v d v= 1 x d x - (v^2+1 )+v v (v^2+1 ) d v= 1 x d x - 1 v d v+ 1 v^2+1 d v= 1 x d x On integrating both sides, we get - v+ ⁻¹ v= x+C ⁻¹ v= x v+C ⁻¹ ( y x )= y+C