question_answer

Solve \[({{x}^{3}}-3x{{y}^{2}})\,dx=({{y}^{3}}-3{{x}^{2}}y)\,dy.\]

Answer:

We have, \[({{x}^{3}}-3x{{y}^{2}})dx=({{y}^{3}}-3{{x}^{2}}y)dy\] \[\Rightarrow \]   \[\frac{dy}{dx}=\frac{{{x}^{3}}-3x{{y}^{2}}}{{{y}^{3}}-3{{x}^{2}}y}\] Clearly, the given equation is a homogeneous equation. Put       y = vx and \[\frac{dy}{dx}=v+x\frac{dv}{dx},\] we get             \[v+x\frac{dv}{dx}=\frac{{{x}^{3}}-3{{x}^{3}}{{v}^{2}}}{{{v}^{3}}{{x}^{3}}-3{{x}^{3}}v}\] \[\Rightarrow \]   \[v+x\frac{dv}{dx}=\frac{1-3{{v}^{2}}}{{{v}^{3}}-3v}\]      \[\Rightarrow \]   \[x\frac{dv}{dx}=\frac{1-3{{v}^{2}}}{{{v}^{3}}-3v}-v=\frac{1-{{v}^{4}}}{{{v}^{3}}-3v}\] \[\Rightarrow \]   \[\int{\frac{{{v}^{3}}-3v}{1-{{v}^{4}}}dv=\int{\frac{1}{x}\,dx}}\] \[\Rightarrow \]  \[\int{\frac{{{v}^{3}}}{1-{{v}^{4}}}dv-3\int{\frac{1}{1-{{v}^{4}}}\,dv=\log |x|}}+\log C\] \[\Rightarrow \]   \[\frac{-\,1}{4}\int{\frac{-\,4{{v}^{3}}}{1-{{v}^{4}}}dv}-\,\frac{3}{2}\int{\frac{2v}{1-{{({{v}^{2}})}^{2}}}\,}dv=\log |Cx|\] \[\Rightarrow \]   \[-\frac{1}{4}\int{\frac{d{{t}_{1}}}{{{t}_{1}}}dv}-\,\frac{3}{2}\int{\frac{d{{t}_{2}}}{1-{{({{t}_{2}})}^{2}}}}=\log |Cx|\] where, \[{{t}_{1}}={{v}^{4}}\] in first integral and \[{{t}_{2}}={{v}^{2}}\] in second integral. \[\Rightarrow \]\[-\frac{1}{4}\log |1-{{v}^{4}}|-\frac{3}{2}\times \frac{1}{2\times 1}\log \left| \frac{1+t}{1-t} \right|=\log |Cx|\] \[\Rightarrow \] \[-\frac{1}{4}\log |1-{{v}^{4}}|-\frac{3}{4}\log \left| \frac{1+{{v}^{2}}}{1-{{v}^{2}}} \right|=\log |Cx|\] \[\Rightarrow \]   \[-\log |1-{{v}^{4}}|-\,3\log \left| \frac{1+{{v}^{2}}}{1-{{v}^{2}}} \right|=4\log |Cx|\] \[\Rightarrow \]   \[\log \left| {{(1-{{v}^{4}})}^{-1}}{{\left( \frac{1+{{v}^{2}}}{1-{{v}^{2}}} \right)}^{-3}} \right|=\log |{{(Cx)}^{4}}|\] \[\Rightarrow \]   \[\frac{1}{1-{{v}^{4}}}\times {{\left( \frac{1+{{v}^{2}}}{1-{{v}^{2}}} \right)}^{3}}={{(Cx)}^{4}}\] \[\Rightarrow \]   \[{{(1-{{v}^{2}})}^{2}}={{(1+{{v}^{2}})}^{4}}{{(Cx)}^{4}}\] \[\Rightarrow \]   \[1-{{v}^{2}}={{(1+{{v}^{2}})}^{2}}{{(Cx)}^{2}}\] \[\Rightarrow \]   \[1-\frac{{{y}^{2}}}{{{x}^{2}}}={{\left( 1+\frac{{{y}^{2}}}{{{x}^{2}}} \right)}^{2}}{{(Cx)}^{2}}\]        \[[\because \,\,v=y/x]\] \[\Rightarrow \]   \[{{x}^{2}}-{{y}^{2}}={{({{x}^{2}}+{{y}^{2}})}^{2}}{{C}^{2}}\]