Question

Variable separable form:
dy/dx=x(2logx+1)/siny+ycosy​

Answer 1

Answer:

$ysiny=x^{2} .logx+c$

Step-by-step explanation:

Variable separable form:

$dy/dx=x(2logx+1)/siny+ycosy$

$(siny +ycosy)dy= [x(2logx+1)]dx$

Now, we need to integrate both the sides.

→$\int\limits sin y.dy +\int\limits cos y.dy$

→ $\int\limits 2x . logxdx +\int\limits xdx$

→ $-cosx y + ysiny + cosy =x^{2} logx-\frac{x^{2} }{2} +\frac{x^{2} }{2} +c$

→ $ysiny=x^{2} .logx+c$

Hope helps!