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Question

x2+y2dydx=8x2-3xy+2y2

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Solution

We have,x2+y2dydx=8 x2-3xy+2y2dydx=8 x2-3xy+2y2x2+y2This is a homogeneous differential equation.Putting y=vx and dydx=v+xdvdx, we getv+xdvdx=8 x2-3vx2+2v2x2x2+v2x2xdvdx=8-3v+2v21+v2-vxdvdx=8-4v+2v2-v31+v2xdvdx=42-v+v22-v1+v2xdvdx=4+v22-v1+v21+v24+v22-vdv=1xdxIntegrating both sides, we get 1+v24+v22-vdv=1xdx .....(1)Let us consider the left hand side of (1).Using partial fraction,Let 1+v24+v22-v=Av+B4+v2+C2-v1+v2=Av2-v+B2-v+C 4+v21+v2=2Av-Av2+2B-Bv+4C+Cv2Comparing the coefficients of both sides, we get 2A-B=0 -A+C=1 & 2B+4C=1Solving these three equations, we getA=-38, B=-34and C=58 1+v24+v22-v=-38v-344+v2+582-v .....(2)From (1) and (2), we get-38v-344+v2+582-v =1xdx -38vv2+4dv-341v2+4dv+5812-vdv=1xdx-316log v2+4-34×2tan -1v2-58log 2-v=log x+log C-34×2tan -1v2=log Cx2-v58v2+4316e-38tan -1v2=Cx2-v58v2+4316Putting v=yx, we gete-38tan -1y2x=Cx2-yx58yx22+4316e-38tan -1y2x=Cx×1x2x-y58y2+4x2316e-38tan -1y2x=C2x-y58y2+4x2316Hence, e-38tan -1y2x=C2x-y58y2+4x2316 is the required solution.

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