Answer to Question #149656 in Differential Equations for Harsh mishra

Question

2x(y+z^2)p+(2y+z^2)q=z^3

Answer 1

$\displaystyle 2x(y+z^2)p+(2y+z^2)q=z^3\\ \textsf{The lagrange auxiliary equation is}\\ \frac{\mathrm{d}x}{2x(y + z^2)} = \frac{\mathrm{d}y}{2y + z^2} = \frac{\mathrm{d}z}{z^3}\\ \textsf{Choosing multipliers}\,\,\left(\frac{1}{x}, -1, -\frac{1}{z}\right)\\ \int\frac{\mathrm{d}x}{x} - \int\mathrm{d}y - \int\frac{\mathrm{d}z}{z} = 0\\ \int\frac{\mathrm{d}x}{x} - \int\mathrm{d}y = \int\frac{\mathrm{d}z}{z} \\ \ln{x} - y - C = \ln{z}\\ \ln\left(\frac{x}{z}\right) = y + C\\ \frac{x}{z} = Ae^{y}, x = Aze^{y}\\ \textsf{Comparing the last two equations}\\ \frac{\mathrm{d}y}{2y + z^2} = \frac{\mathrm{d}z}{z^3}\\ \frac{\mathrm{d}y}{\mathrm{d}z} =\frac{2y + z^2}{z^3}\\ \frac{\mathrm{d}y}{\mathrm{d}z} =\frac{2y}{z^3} + \frac{1}{z}\\ \frac{\mathrm{d}y}{\mathrm{d}z} - \frac{2y}{z^3} = \frac{1}{z}\\ \textsf{Integration Factor}\,(IF) = e^{-\int \frac{2}{z^3}\, \mathrm{d}z} = e^{\frac{1}{z^2}}\\ y\cdot IF = \int \frac{IF}{z}\,\mathrm{d}z\\ ye^{\frac{1}{z^2}} = \int \frac{e^{\frac{1}{z^2}}}{z} \,\mathrm{d}z\\ \textsf{Let}\,\, u = \frac{1}{z^2}, -\frac{z^3}{2}\mathrm{d}u = \mathrm{d}z\\ ye^{\frac{1}{z^2}} = \int \frac{e^{u}}{z} \cdot -\frac{z^3}{2}\mathrm{d}u \\ ye^{\frac{1}{z^2}} = \int e^u\cdot -\frac{z^2}{2}\mathrm{d}u \\ ye^{\frac{1}{z^2}} = \frac{-1}{2}\int \frac{e^u}{u}\,\mathrm{d}u \\ \textsf{The integral on the right side}\\ \textsf{can be written in terms}\\ \textsf{of the the exponential integral}\,\, \mathrm{Ei}(u)\\ ye^{\frac{1}{z^2}} = \frac{-1}{2}\int_0^u \frac{e^t}{t}\,\mathrm{d}t \\ ye^{\frac{1}{z^2}} = \frac{-\mathrm{Ei}(u)}{2}\,\, y = \frac{-e^{\frac{-1}{z^2}}\mathrm{Ei}(u)}{2}\\ y = \frac{-e^{\frac{-1}{z^2}}\mathrm{Ei}\left(\frac{1}{z^2}\right)}{2}\\ \textsf{Therefore, the solution to PDE is}\,\,\\ x = Aze^{y}, y = \frac{-e^{\frac{-1}{z^2}}\mathrm{Ei}\left(\frac{1}{z^2}\right)}{2}\\$