If a curve passes through the origin and the slope of the tangent to it at any point x,y is  x2−4x+y+8x−2 then this curve also passes through the point

dydx=x2-4x+y+8x-2=(x-2)2+(y+4)(x-2)dydx=(x-2)+(y+4)(x-2)∣ plug in x-2=t⇒dx=dty+4=u⇒dy=du,dydx=dudt⇒dudt=t+ut⇒dudt-ut=t I.F =e∫-1tdt=e-ln(t)=1t Solution is, u·1t=∫t·1tdt⇒ut=t+cy+4x-2=x-2+c Passing through (0,0)⇒c=0⇒  y+4=(x-2)2