The solution of the equation (d y/d x)=(x(2 log x+1)/ sin -y+y cos y)

Question

The solution of the equation (d y/d x)=(x(2 log x+1)/ sin -y+y cos y) (A) y sin y=x2 log x+(x2/y)+c (B) y cos y=x2( log x+1)+c (C) y cos y=x2 log x+(x2/2)+c (D) y sin y=x2 log x+c

Tardigrade Admin · Accepted Answer

(y cos y+ sin y) d y=(2 x log x+x) d x y sin y-∫ sin y d y+∫ sin y d y =x2 log x-∫ x2 ⋅ (1/x) d x+∫ x d x+c ∴ y sin y=x2 log x+c

Q. The solution of the equation $\frac{d y}{d x} = \frac{x ( 2 l o g x + 1 )}{s i n - y + y c o s y}$

$y sin y = x^{2} lo g x + \frac{x ^{2}}{y} + c$

$y cos y = x^{2} (lo g x + 1) + c$

$y cos y = x^{2} lo g x + \frac{x ^{2}}{2} + c$

$y sin y = x^{2} lo g x + c$

$(y cos y + sin y) d y = (2 x lo g x + x) d x$
$y sin y - \int sin y d y + \int sin y d y$
$= x^{2} lo g x - \int x^{2} \cdot \frac{1}{x} d x + \int x d x + c$
$∴ y sin y = x^{2} lo g x + c$

Q. The solution of the equation dxdy​=sin−y+ycosyx(2logx+1)​

ysiny=x2logx+yx2​+c

ycosy=x2(logx+1)+c

ycosy=x2logx+2x2​+c

ysiny=x2logx+c