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

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

Posted by Meenakshi Rajput 7 years, 11 months ago

CBSE > Class 12 > Mathematics

1 answers

Sahdev Sharma 7 years, 11 months ago

${dy\over dx}={x(2log\ x+1)\over sin\ y+y.cos\ y}$

$(sin \ y+y.cos\ y)dy = [x(2log\ x+1)]dx$

Integrating both sides, we get

$\int sin\ y\ dy +\int y.cos\ y\ dy = \int 2x.log\ x \ dx+ \int xdx$

$-cox \ y + y\ sin\ y + cos \ y = x^2log\ x -{x^2\over 2}+ {x^2\over 2}+C$

$=> y sin\ y =x^2.log \ x+ C$

7Thank You

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