The solution of the differential equation (ydx - xdy/xy)=xdx+ydy is (where, C is an arbitrary constant)

Question

The solution of the differential equation (ydx - xdy/xy)=xdx+ydy is (where, C is an arbitrary constant) (A) (x/y)=x+y+C (B) (x/y)=(x2 + y2/2)+C (C) ln ((x/y))=x2+y2+C (D) 2ln((x/y))=x2+y2+C

Tardigrade Admin · Accepted Answer

The given equation is (1/((x)y))⋅ (ydx - xdy/y2)=xdx+ydy or d ( ln (( x / y )))= x d x + ydy On integrating, we get, ∫ d ( ln (( x / y )))=∫ xdx +∫ ydy ⇒ ln (( x / y ))=( x 2/2)+( y 2/2)+ k ⇒ 2 ln (( x / y ))= x 2+ y 2+ C

Q. The solution of the differential equation $\frac{y d x - x d y}{x y} = x d x + y d y$ is (where, $C$ is an arbitrary constant)

$\frac{x}{y} = x + y + C$

$\frac{x}{y} = \frac{x ^{2} + y ^{2}}{2} + C$

$l n (\frac{x}{y}) = x^{2} + y^{2} + C$

$2 l n (\frac{x}{y}) = x^{2} + y^{2} + C$

Q. The solution of the differential equation xyydx−xdy​=xdx+ydy is (where, C is an arbitrary constant)

yx​=x+y+C

yx​=2x2+y2​+C

ln(yx​)=x2+y2+C

2ln(yx​)=x2+y2+C

The given equation is (yx​)1​⋅y2ydx−xdy​=xdx+ydy or d(ln(yx​))=xdx+ydy On integrating, we get, ∫d(ln(yx​))=∫xdx+∫ydy ⇒ln(yx​)=2x2​+2y2​+k ⇒2ln(yx​)=x2+y2+C

Q. The solution of the differential equation $\frac{y d x - x d y}{x y} = x d x + y d y$ is (where, $C$ is an arbitrary constant)

$\frac{x}{y} = x + y + C$

$\frac{x}{y} = \frac{x ^{2} + y ^{2}}{2} + C$

$l n (\frac{x}{y}) = x^{2} + y^{2} + C$

$2 l n (\frac{x}{y}) = x^{2} + y^{2} + C$