Question

Find the particular solution of the differential equation
y2+ x 2 dy/dx = x y dy/dx satisfying the given condition y=1 when x=1

Answer 1

Dear Student,


$$\begin{gathered} \mathrm{Given}, \\ {\mathrm{y}}^2+{\mathrm{x}}^2\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{xy}\frac{\mathrm{dy}}{\mathrm{dx}} \\ {\mathrm{y}}^2=\frac{\mathrm{dy}}{\mathrm{dx}}\left(\mathrm{xy}-{\mathrm{x}}^2\right) \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{y}}^2}{\left(\mathrm{xy}-{\mathrm{y}}^2\right)} \\ \mathrm{y}=\mathrm{vx} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\frac{\mathrm{xdv}}{\mathrm{dx}} \\ \mathrm{v}+\frac{\mathrm{xdv}}{\mathrm{dx}}=\frac{{\mathrm{v}}^2{\mathrm{x}}^2}{{\mathrm{vx}}^2-{\mathrm{v}}^2{\mathrm{x}}^2} \\ \mathrm{v}+\frac{\mathrm{xdv}}{\mathrm{dx}}=\frac{\mathrm{v}}{1-\mathrm{v}} \\ \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}}{1-\mathrm{v}}-\mathrm{v} \\ \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{\mathrm{v}-\mathrm{v}+{\mathrm{v}}^2}{1-\mathrm{v}} \\ \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{{\mathrm{v}}^2}{1-\mathrm{v}} \\ \left(\frac{1-\mathrm{v}}{{\mathrm{v}}^2}\right)\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}} \\ \left(\frac{1}{{\mathrm{v}}^2}-\frac{1}{\mathrm{v}}\right)\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}} \\ \mathrm{integrating}\mathrm{both}\mathrm{sides}: \\ \int \left(\frac{1}{{\mathrm{v}}^2}-\frac{1}{\mathrm{v}}\right)\mathrm{dv}=\int \frac{\mathrm{dx}}{\mathrm{x}} \\ =\frac{-1}{\mathrm{v}}-\ln \left|\mathrm{v}\right|=\ln \left|\mathrm{x}\right|+\mathrm{C} \\ \mathrm{put}\mathrm{v}=\frac{\mathrm{y}}{\mathrm{x}} \\ =\frac{\mathbf{-}\mathbf{x}}{\mathbf{y}}\mathbf{-}\mathbf{ln}\left|\frac{\mathbf{y}}{\mathbf{x}}\right|\mathbf{=}\mathbf{ln}\left|\mathbf{x}\right|\mathbf{+}\mathbf{C} \end{gathered}$$



Regards!