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]))