Solve the differential equation \((x^{2}-3y^{2})\,dx+2xy\,dy=0\).
Step 1 Both \(x^{2}-3y^{2}\) and \(2xy\) are homogeneous of degree 2. Do you see why?
Step 2 Let \(y=xv\). Then \(dy=x\,dv+v\,dx\). Substitute these into the differential equation: \[ \begin{eqnarray*} (x^{2}-3y^{2})\,dx+2xy\,dy& =&0 \\ [ x^{2}-3( xv) ^{2}]\,dx+2x (xv) (x\,dv+v\,dx)& =&0\qquad \color{#0066A7}{{dy=x\,} {dv=} {+} {v} {dx}} \\ ( x^{2}-x^{2}v^{2})\,dx+2x^{3}v\,dv& =&0 \\ x^{2}(1-v^{2})\,dx+2x^{3}v\,dv& =&0 \end{eqnarray*} \]
Step 3 Then for \(x\neq 0\) and \(v\neq \pm 1,\) we can separate the variables by dividing by \(x^{3}\) and \(1-v^{2}\). \[ \begin{eqnarray*} x^{2} ( 1-v^{2})\,dx+2x^{3}v\,dv &=&0 \\ \hspace{-2pc}\dfrac{x^{2}( 1-v^{2})\,dx}{x^{3}( 1-v^{2}) }+\dfrac{2x^{3}v\,dv}{x^{3}( 1-v^{2}) } &=&0 \\ \dfrac{dx}{x}+\dfrac{2v\,dv}{1-v^{2}} &=&0 \end{eqnarray*} \]
Step 4 Integrate. \[ \begin{eqnarray*} \hspace{-5pc}\int \frac{dx}{x}-\int \frac{2v\,dv}{v^{2}-1} &=&0\\ \ln \left\vert x\right\vert -\ln \left\vert v^{2}-1\right\vert &=&C_{1} \\ \ln \left\vert \dfrac{x}{v^{2}-1}\right\vert &=&C_{1} \end{eqnarray*} \]
Since \(C_{1}\) is a constant, we can write \(C_{1}=\ln C_{2},\) where \(C_{2}\) is a constant. Then \[ \begin{eqnarray*} \ln \left\vert \dfrac{x}{v^{2}-1}\right\vert &=&\ln C_{2} \\ \left\vert \dfrac{x}{v^{2}-1}\right\vert &=&C_{2} \\ \left\vert \dfrac{x}{\dfrac{y^{2}}{x^{2}}-1}\right\vert &=&C_{2}\qquad \color{#0066A7}{{v}} \color{#0066A7}{\hbox{\(=\dfrac{y}{x}\)}}\\ \left\vert \dfrac{x^{3}}{y^{2}-x^{2}}\right\vert & =&C_{2} \\ x^{3}& =&\pm C_{2}( y^{2}-x^{2}) \end{eqnarray*} \]
Since \(\pm C_{2}\) may be either positive or negative, the general solution can be written as \(x^{3}=C( y^{2}-x^{2}) ,\) where \(C\neq 0\) is a constant.