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Question:

The solution of the differential equation dydx + y²secx = tanx/2y, where 0 ≤ x < π/2 and y(0) = 1, is given by:

y²=1-xsecx+tanx

y²=1+xsecx+tanx

y=1+xsecx+tanx

y=1-xsecx+tanx

Solution:

dydx+y²secx=tanx2y2ydydx+y²secx=tanxPut y²=t ⇒2ydydx=dtdxdtdx+tsecx=tanxI.f=e∫secxdx=eln(secx+tanx)=secx+tanxdtdx(secx+tanx)+tsecx(secx+tanx)=tanx(secx+tanx)∫d(t(secx+tanx))=tanx(secx+tanx)dxt(secx+tanx)=secx+tanx-xt=1-xsecx+tanx, y²=1-xsecx+tanx