Solution of the differential equations cosxdy=y(sinx-y)dx,0<x<π2 is
secx=(tanx+C)y
ysecx=tanx+C
ytanx=secx+C
tanx=(secx+C)y
Explanation for the correct option:
Simplifying and Integrating the given expression:
cosxdy=y(sinx-y)dxdydx=y(sinx-y)cosxsecxdydx=ytanx-y2secx∵sinθcosθ=tanθand1cosθ=secθdydx=y21ytanx-secx1y2dydx-1ytanx=-secx
Substituting
1y=t∴1y2dydx=dtdx
-dtdx-ttanx=-secxdtdx+ttanx=secx
Now, Integrating the above expression we get
IF=etanxdx=secxwhereIF=integratingfactor
Solution is
t(IF)=∫(IF)secxdx=1ysecx=tanx+c
Therefore, option (A) is the correct answer.
Find a particular solution of the differential equation, given that y = – 1, when x = 0 (Hint: put x – y = t)