Question

Solve the differential equation (2x+y)dx+(x−2y)dy = 0

Answer 1

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))