The solution to the differential equation y log y+xy'=0, where y (1) = e, is

Question

The solution to the differential equation y log y+xy'=0, where y (1) = e, is (A) x(log y)=1 (B) xy(log y)=1 (C) (log y)2=2 (D) log y+((x2/2))y=1

Tardigrade Admin · Accepted Answer

x (dy/dx)+y (log y)=0 or ∫ (dx/x)+∫ (dy/y(log y))=c or log x+log(log y)=log c or x log y=c y (1)=e ⇒ c=1 Hence, the equation of the curve is x log y = 1

Q. The solution to the differential equation $y l o g y + x y^{'} = 0$ , where $y (1) = e$ , is

$x (l o g y) = 1$

$x y (l o g y) = 1$

$(l o g y)^{2} = 2$

$l o g y + (\frac{x ^{2}}{2}) y = 1$

$x \frac{d y}{d x} + y (l o g y) = 0$
or $\int \frac{d x}{x} + \int \frac{d y}{y ( l o g y )} = c$
or $l o g x + l o g (l o g y) = l o g c$
or $x l o g y = c$
$y (1) = e$
$\Rightarrow c = 1$
Hence, the equation of the curve is $x l o g y = 1$

Q. The solution to the differential equation ylogy+xy′=0, where y(1)=e, is

x(logy)=1

xy(logy)=1

(logy)2=2

logy+(2x2​)y=1