Answer to Question #149012 in Differential Equations for Nikhil

Question

Solve the differential equation
dy/dx= x+2y-3/2x+y-3

Answer 1

$\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{x + 2y - 3}{2x + y - 3}\\ (2x + y - 3)\mathrm{d}y + (x + 2y - 3)\mathrm{d}x = 0\\ M\mathrm{d}y + N\mathrm{d}x = 0\\ M_x = \frac{\partial M}{\partial x} = 2\\ N_x = \frac{\partial N}{\partial y} = 2\\ \textsf{Since}\,\, M_y = N_x\,\,\textsf{the differential equation is exact}\\ \frac{\partial z}{\partial y} = 2x + y - 3\\ z = \int(2x + y - 3)\, \mathrm{d}y\\ z = 2xy + \frac{y^2}{2} - 3y + f(x)\\ \frac{\partial z}{\partial x} = x + 2y - 3\\ z = \int (x + 2y - 3)\,\mathrm{d}x\\ z = \frac{x^2}{2} + 2xy - 3x + f(y)\\ z = 2xy + \frac{x^2 + y^2}{2} - 3(x + y)$