Question Number 100594 by Coronavirus last updated on 27/Jun/20

solve the differential equations 1- xcos (ln (x/y))dy−ydx=0 2- ydx+2xdy =2y((√x)/(cos^2 (y)))dy y(0)=π

Answered by smridha last updated on 27/Jun/20

2.[(dx/(2(√x)))+((√x)/y)dy=sec^2 (y)dy] mult: by y both sides and integrating ∫d(y.(√x))=∫y.sec^2 (y)dy ⇒y(√x)=ytan(y)+ln[cos(y)]+c put the condition y(0)=𝛑 ⇒0=0+ln(−1)+c so c=+_− i𝛑 so the solution y(√x)=ytan(y)+ln[cos(y)]+_− i𝛑

Answered by smridha last updated on 28/Jun/20

(1).(dx/dy)=(x/(2y))[((x/y))^i +((x/y))^(−i) ] let (x/y)=v so (dx/dy)=v+y.(dv/dy) now v+y.(dv/dy)=(v/2)[((v^(2i) +1)/v^i )] y.(dv/dy)=v[(((v^i −1)^2 )/(2v^i ))] (dy/y)=((2v^(i−1) )/((v^i −1)^2 ))dv integrating both sides we get.. ln(y)=2i.(1/((v^i −1)))+ln(c) so y=c.e^((2i)/([((x/y))^i −1]))