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Question:

Find the particular solution of the differential equation dydx=x(2logx+1)siny+ycosy given that y=π/2 when x=1.

Solution:

dydx=x(2logx+1)siny+ycosy⇒(siny+ycosy)dy=x(2logx+1)dx⇒∫(siny+ycosy)dy=∫x(2logx+1)dx⇒∫sinydy+∫ycosydy=∫2xlogxdx+∫xdx⇒−cosy+ysiny+cosy=2(12x2ln(x)−x24)+x22+C⇒ysiny=(x2ln(x)−x22)+x22+C⇒ysiny=x2ln(x)+CThe required solution isx2ln(x)−ysiny+C=0.When(x,y)≡(1,π2),We getC=π2Hence, the particular solution isx2ln(x)−ysiny+π2=0.