Given, the differential equation(2x−y)dx+(4x+y−6)dy=0Toremove−6,wesubstitutey=z+a,x=t+bdtdy=dtdzanddtdx=1dxdy=dtdzNow, finding the value of a,b.By comparing the coefficients ,(2x−y)=0⟹2(t+b)−(z+a)=0And,(4x+y−6)=0⟹4(t+b)+(z+a)−6=0Solve for a and b,a=2,b=1Then, the given differential equations,(2(t+1)−(z+2))+(4(t+1)+(z+2)−6)dtdz=0(2t−z)+(4t+z)dtdz=0dtdz=z+4tz−2tThis is a homogenous differential equation.Putz=vt⟹dtdz=v+tdtdvThen,z+4tz−2t=v+tdtdvv+tdtdv=vt+4tvt−2tTaking t outside in right hand side and simplify, v+tdtdv=v+4v−2tdtdv=v+4v−2−vsimplify right hand side,tdtdv=(−1)v+4v2+3v+2v2+3v+2v+4dv=(−1)tdtBy multiply and divide by 2 and then split the numerator,21(v2+3v+22v+3+v2+3v+25)dv=(−1)tdtIntegrating both side,21∫(v2+3v+22v+3+v2+3v+25)dv=(−1)∫tdt21(∫v2+3v+22v+3dv+∫v2+3v+25dv)=(−1)∫tdtBy using partial fraction of second term of left hand side,21(∫v2+3v+22v+3dv+5∫(v+11−v+21)dv)=(−1)∫tdtBy using integration formula,∫f(x)f′(x)dx=ln(f(x)),21(ln(v2+3v+2)+5(ln(v+1)−ln(v+2))=(−1)lnt+lnC21(ln(v2+3v+2)+5lnv+2v+1)=lntC21(ln(v2+3v+2)+5lnv+2v+1)=lntC21(ln(v+1)(v+2)+5lnv+2v+1))=lntC21(ln((v+2)5(v+1)(v+2)(v+1)5))=lntCln(v+2)4(v+1)6=lntC(v+2)2(v+1)3=tCPut v=tz,(tz+2)2(tz+1)3=tCPut z=y−2,t=x−1(x−1y−2+2)2(x−1y−2+1)3=x−1C(2x+y−4)2(x+y−3)3=CThis is the required solution.
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