A solution of differential equation (sec2y)dydx+2x tan y=x3, is
2 tan y=c.e−x2+x2−1
tan y=c.e−x2+x2−1
tan y=cex2+x2−1
None of these
tan y=z⇒sec2 ydydx=dzdxdzdx+2xz=x3 I.F=e∫2xdx=ex2 Solution is z.ex2=∫x3.ex2dx Let t=x2⇒dt=2x dx⇒tan y.ex2=12[t.et−et]+c⇒2 tan y=ce−x2+x2−1