Answer:
y(x) = 1/2 (x - sqrt(5 x^2 + c_1)) or y(x) = 1/2 (x + sqrt(5 x^2 + c_1))
Step-by-step explanation:
Solve 2 x + y(x) + (dy(x))/(dx) (x - 2 y(x)) = 0:
Let P(x, y) = 2 x + y and Q(x, y) = x - 2 y.
This is an exact equation, because (dP(x, y))/(dy) = 1 = (dQ(x, y))/(dx).
Define f(x, y) such that (df(x, y))/(dx) = P(x, y) and (df(x, y))/(dy) = Q(x, y).
Then, the solution will be given by f(x, y) = c_1, where c_1 is an arbitrary constant.
Integrate (df(x, y))/(dx) with respect to x in order to find f(x, y):
f(x, y) = integral(2 x + y) dx = x^2 + x y + g(y) where g(y) is an arbitrary function of y.
Differentiate f(x, y) with respect to y in order to find g(y):
(df(x, y))/(dy) = d/(dy) (x^2 + y x + g(y)) = x + (dg(y))/(dy)
Substitute into (df(x, y))/(dy) = Q(x, y):
x + (dg(y))/(dy) = x - 2 y
Solve for (dg(y))/(dy):
(dg(y))/(dy) = -2 y
Integrate (dg(y))/(dy) with respect to y:
g(y) = integral-2 y dy = -y^2
Substitute g(y) into f(x, y):
f(x, y) = x^2 - y^2 + y x
The solution is f(x, y) = c_1:
x^2 - y^2 + y x = c_1
Solve for y:
y(x) = 1/2 (x - sqrt(5 x^2 - 4 c_1)) or y(x) = 1/2 (x + sqrt(5 x^2 - 4 c_1))
Simplify the arbitrary constants:
Answer: y(x) = 1/2 (x - sqrt(5 x^2 + c_1)) or y(x) = 1/2 (x + sqrt(5 x^2 + c_1))
