Answer to Question #190503 in Differential Equations for randz
2. (2x − y)dx + (4x + y − 6)dy = 0
1.
"\\text{Given, the differential equation} (2x-y)dx+(4x+y-6)dy=0\\newline\nTo remove -6, we substitute y=z+a, x=t+b\\newline\n\\frac{dy}{dt}=\\frac{dz}{dt} and \\frac{dx}{dt}=1 \\newline\n\\frac{dy}{dx}=\\frac{dz}{dt}\\newline\n\\text{Now, finding the value of a,b.}\\newline\n\\text{By comparing the coefficients ,}\\newline\n(2x-y)=0\\newline\n\\implies2(t+b)-(z+a)=0\\newline\n\\text{And,} (4x+y-6)=0\\implies4(t+b)+(z+a)-6=0\\newline\n\\text{Solve for a and b,}\\newline\na=2,b=1\\newline\n\\text{Then, the given differential equations,}\\newline\n(2(t+1)-(z+2))+(4(t+1)+(z+2)-6)\\frac{dz}{dt}=0\\newline\n(2t-z)+(4t+z)\\frac{dz}{dt}=0\\newline\n\\frac{dz}{dt}=\\frac{z-2t}{z+4t}\\newline\n\\text{This is a homogenous differential equation.}\\newline\n\\text{Put} z=vt\\implies\\frac{dz}{dt}=v+t\\frac{dv}{dt}\\newline\n\\text{Then,}\\newline\n\\frac{z-2t}{z+4t}=v+t \\frac{dv}{dt}\\newline\nv+t \\frac{dv}{dt}=\\frac{vt-2t}{vt+4t}\\newline\n\\text{Taking t outside in right hand side and simplify, }\\newline\nv+t \\frac{dv}{dt}=\\frac{v-2}{v+4}\\newline\nt\\frac{dv}{dt}=\\frac{v-2}{v+4}-v\\newline\n\\text{simplify right hand side,}\\newline\nt\\frac{dv}{dt}=(-1)\\frac{v^2+3v+2}{v+4}\\newline\n\\frac{v+4}{v^2+3v+2}dv=(-1)\\frac{dt}{t}\\newline\n\\text{By multiply and divide by 2 and then split the numerator,}\\newline\n\\frac{1}{2}(\\frac{2v+3}{v^2+3v+2}+\\frac{5}{v^2+3v+2})dv=(-1)\\frac{dt}{t}\\newline\n\\text{Integrating both side,}\\newline\n\\frac{1}{2}\\int(\\frac{2v+3}{v^2+3v+2}+\\frac{5}{v^2+3v+2})dv=(-1)\\int\\frac{dt}{t}\\newline\n\\frac{1}{2}(\\int\\frac{2v+3}{v^2+3v+2}dv+\\int\\frac{5}{v^2+3v+2}dv)=(-1)\\int\\frac{dt}{t}\\newline\n\\text{By using partial fraction of second term of left hand side,}\\newline\n\\frac{1}{2}(\\int\\frac{2v+3}{v^2+3v+2}dv+5\\int(\\frac{1}{v+1}-\\frac{1}{v+2})dv)=(-1)\\int\\frac{dt}{t}\\newline\n\\text{By using integration formula,} \\int \\frac{f'(x)}{f(x)}dx=ln(f(x)),\\newline\n\\frac{1}{2}(ln(v^2+3v+2)+5(ln(v+1)-ln(v+2))=(-1)lnt+lnC\\newline\n\\frac{1}{2}(ln(v^2+3v+2)+5ln\\frac{v+1}{v+2})=ln\\frac{C}{t}\\newline\n\\frac{1}{2}(ln(v^2+3v+2)+5ln\\frac{v+1}{v+2})=ln\\frac{C}{t}\\newline\n\\frac{1}{2}(ln(v+1)(v+2)+5ln\\frac{v+1}{v+2}))=ln\\frac{C}{t}\\newline\n\\frac{1}{2}(ln(\\frac{(v+1)(v+2)(v+1)^5}{(v+2)^5}))=ln\\frac{C}{t}\\newline\nln\\sqrt\\frac{(v+1)^6}{(v+2)^4}=ln\\frac{C}{t}\\newline\n\\frac{(v+1)^3}{(v+2)^2}=\\frac{C}{t}\\newline\n\\text{Put }v=\\frac{z}{t},\\newline\n\\frac{(\\frac{z}{t}+1)^3}{(\\frac{z}{t}+2)^2}=\\frac{C}{t}\\newline\n\\text{Put }z=y-2, t=x-1\\newline\n\\frac{(\\frac{y-2}{x-1}+1)^3}{(\\frac{y-2}{x-1}+2)^2}=\\frac{C}{x-1}\\newline\n\\frac{(x+y-3)^3}{(2x+y-4)^2}=C\\newline\n\n\\text{This is the required solution.}"
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