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(a) Let, Mdx+Ndy=0 has an intograting factor{:[mu(x","y).],[:.mu(x","y)M(x","y)dx+mu(x","y)N(x","y)dy=0" is exact. "],[=>(mu M)_(y)=(mu N)x],[=>mu_(y)M+mu My=mu_(x)N+mu Nx],[=>mu_(y)M-mu_(x)N=mu(N_(x)-My)dots]:}If we write mu(x,y)=g(x+y)then mu_(x)=g^(')(x+y)mu_(y)=g^(')(x+y)from, (x) g^(')(x+y)M-g^(')(x+y)N=g(x+y)(N_(x)-M_(y))=>quad(g^(')(x+y))/(g(x+y))=(N_(x)-M_(y))/(M-N^(2))This shows that, ((del N)/(del x)-(del M)/(del y))/(M-N) depends onty on (x+y) and converse.Also, quad ln(g(x+y))=int((N_(x)-M_(y))/(M-N))d(x+y){:[=>g(x+y)=e^(int((N_(x)-My)/(M-N))d(x+y))],[:.quad M(x","y)=e^(int(N_(x)-My)/(M-N)d(x+y))]:}Which satisfies the requirements of an integrating factor.(b){:[(3+y+xy)dx+(3 ... See the full answer