solve, (xy+2x+y+2)dx + (x^2 +2x)dy =0.
Answers
Answer:
Simplifying
(xy + 2x + y + 2) * dx + (x2 + 2x) * dy = 0
Reorder the terms:
(2 + 2x + xy + y) * dx + (x2 + 2x) * dy = 0
Reorder the terms for easier multiplication:
dx(2 + 2x + xy + y) + (x2 + 2x) * dy = 0
(2 * dx + 2x * dx + xy * dx + y * dx) + (x2 + 2x) * dy = 0
Reorder the terms:
(2dx + dxy + 2dx2 + dx2y) + (x2 + 2x) * dy = 0
(2dx + dxy + 2dx2 + dx2y) + (x2 + 2x) * dy = 0
Reorder the terms:
2dx + dxy + 2dx2 + dx2y + (2x + x2) * dy = 0
Reorder the terms for easier multiplication:
2dx + dxy + 2dx2 + dx2y + dy(2x + x2) = 0
2dx + dxy + 2dx2 + dx2y + (2x * dy + x2 * dy) = 0
2dx + dxy + 2dx2 + dx2y + (2dxy + dx2y) = 0
Reorder the terms:
2dx + dxy + 2dxy + 2dx2 + dx2y + dx2y = 0
Combine like terms: dxy + 2dxy = 3dxy
2dx + 3dxy + 2dx2 + dx2y + dx2y = 0
Combine like terms: dx2y + dx2y = 2dx2y
2dx + 3dxy + 2dx2 + 2dx2y = 0
Solving
2dx + 3dxy + 2dx2 + 2dx2y = 0
Solving for variable 'd'.
Move all terms containing d to the left, all other terms to the right.
Factor out the Greatest Common Factor (GCF), 'dx'.
dx(2 + 3y + 2x + 2xy) = 0