Problem 16 - Entrance Test

(x² - yx²) dy + (y² + xy²) dx = 0 => x²(1 - y) dy + y²(1 + x) dx = 0 => x²(1 - y) dy = -y²(1 + x) dx => (1 - y)/y² dy = -(1 + x)/x² dx => (1/y² - 1/y) dy = -(1/x² + 1/x) dx. Integrating both sides: ∫ (1/y² - 1/y) dy = -∫ (1/x² + 1/x) dx => -1/y - ln|y| = -(-1/x + ln|x|) + c => -1/y - ln|y| = 1/x - ln|x| + c => ln|x| - ln|y| = 1/x + 1/y + c => ln|x/y| = (x+y)/xy + c => ln|x/y| - (x+y)/xy = c. Exponentiating both sides is very hard. Rearranging terms : (y+1)y² dx + (1-y)x² dy = 0. (1+x)/x² dx + (1-y)/y² dy = 0, intergrating gives -1/x + ln(x) -1/y - ln(y) = C. Rearrange gives -1/x-1/y + ln(xy) = c. -(x+y)/xy + ln(xy) = c. Then xy Ln(xy) - x -y = C xy. So x+ y = cxy + xy ln xy. None match the answer. Check if separable (1-y)x² dy = (-(1+x))y² dx => (1-y/y²) dy = -((1+x)/x²) dx integral => -1/y - ln(y) = 1/x - ln(x) = > Rearranging gives => ln x /y= 1/x + 1/y = so xy ln(x/y) + x + y = 0. try x + y = cxy if dy/dx = cxy. So it does seem none of these work. The result does NOT match the answers.