If the slope of the tangent at (x,y) to a curve passing through 1,π4 is given by yx-cos2yx , then the equation of the curve is

Given, dydx=yx−cos2yx Putting, y=vx so that dydx=v+xdvdx We get, v+xdvdx=v−cos2v⇒dvcos2v=−dxx⇒sec2vdv=−dxx Integrating, we get tanv=−lnx+lnc                                tanyx=−lnx+lnc This passes through 1,π4⇒lnc=1=loge∴tanyx=-logx+loge=logex⇒y=xtan−1logex