Question

Solve:

( x³ + 3xy² )dx + ( y³ + 3x² y) dy = 0

The answer given in the book is : y⁴ + 6x² y² + x⁴ = C₁

Answer 1

$$\begin{gathered} \left({\mathrm{x}}^3+3{\mathrm{xy}}^2\right)\mathrm{dx}=-\left({\mathrm{y}}^3+3{\mathrm{x}}^2\mathrm{y}\right)\mathrm{dy} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-{\mathrm{x}}^3-3{\mathrm{xy}}^2}{{\mathrm{y}}^3+3{\mathrm{x}}^2\mathrm{y}}.....\left(1\right) \\ \mathrm{Let}\mathrm{y}=\mathrm{vx},\mathrm{on}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get}, \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}} \\ \mathrm{Put}\mathrm{this}\mathrm{value}\mathrm{in}\mathrm{equation}\left(1\right), \\ \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-{\mathrm{x}}^3-3{\mathrm{x}}^3{\mathrm{v}}^2}{{\mathrm{v}}^3{\mathrm{x}}^3+3{\mathrm{vx}}^3} \\ \Rightarrow \mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-1-3{\mathrm{v}}^2}{{\mathrm{v}}^3+3\mathrm{v}} \\ \Rightarrow \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-1-3{\mathrm{v}}^2}{{\mathrm{v}}^3+3\mathrm{v}}-\mathrm{v} \\ \Rightarrow \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-1-3{\mathrm{v}}^2-{\mathrm{v}}^4-3{\mathrm{v}}^2}{{\mathrm{v}}^3+3\mathrm{v}} \\ \Rightarrow \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{-\left(1+6{\mathrm{v}}^2+{\mathrm{v}}^4\right)}{{\mathrm{v}}^3+3\mathrm{v}} \\ \Rightarrow -\frac{{\mathrm{v}}^3+3\mathrm{v}}{\left(1+6{\mathrm{v}}^2+{\mathrm{v}}^4\right)}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}} \\ \mathrm{Integrating}\mathrm{on}\mathrm{both}\mathrm{side}, \\ -\int \frac{{\mathrm{v}}^3+3\mathrm{v}}{\left(1+6{\mathrm{v}}^2+{\mathrm{v}}^4\right)}\mathrm{dv}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\ \Rightarrow -\int \frac{1}{4}\left(\frac{4{\mathrm{v}}^3+12\mathrm{v}}{\left(1+6{\mathrm{v}}^2+{\mathrm{v}}^4\right)}\right)\mathrm{dv}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\ \left(\mathrm{Let}1+6{\mathrm{v}}^2+\mathrm{v}=\mathrm{t}\Rightarrow 4{\mathrm{v}}^3+12\mathrm{v}\mathrm{dv}=\mathrm{dt}\right) \\ \Rightarrow -\frac{1}{4}\int \left(\frac{1}{t}\right)dt=\int \frac{\mathrm{dx}}{\mathrm{x}} \\ \Rightarrow -\frac{1}{4}\log tdt=\log x+\log c \\ \Rightarrow -\frac{1}{4}\log \left(1+6{\mathrm{v}}^2+{\mathrm{v}}^4\right)=\log x+\log c \\ \Rightarrow -\log \left(1+6{\mathrm{v}}^2+{\mathrm{v}}^4\right)=4\log \left(cx\right) \\ \Rightarrow \log {\left(1+6{\mathrm{v}}^2+{\mathrm{v}}^4\right)}^{-1}=\log {\left(cx\right)}^4 \\ \Rightarrow \frac{1}{\left(1+6{\mathrm{v}}^2+{\mathrm{v}}^4\right)}{c}^4{x}^4=1 \\ \Rightarrow \frac{1}{\left(1+6{\left({\displaystyle \frac{y}{x}}\right)}^2+{\left(\frac{y}{x}\right)}^4\right)}{c}^4{x}^4=1 \\ \Rightarrow c^4{x}^4\left(\frac{{\mathrm{y}}^4+6{\mathrm{x}}^2{\mathrm{y}}^2+{\mathrm{x}}^4}{x^4}\right)=1 \\ \Rightarrow {\mathrm{y}}^4+6{\mathrm{x}}^2{\mathrm{y}}^2+{\mathrm{x}}^4=\frac{1}{{\mathrm{c}}^4} \\ \Rightarrow {\mathbf{y}}^{\mathbf{4}}\mathbf{+}\mathbf{6}{\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}\mathbf{+}{\mathbf{x}}^{\mathbf{4}}\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\mathrm{where},{\mathrm{C}}_1=\frac{1}{{\mathrm{c}}^4} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$