Answer to Question #190503 in Differential Equations for randz

Question

2. (2x − y)dx + (4x + y − 6)dy = 0

Answer 1

1.

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