Prove the equation is with linear coefficient. Solve the following

Question

(-3x+y-1)dx + (x+y+3)dy = 0

Answer 1

Solve -1 - 3 x + y(x) + (dy(x))/(dx) (x + y(x) + 3) = 0:

Let P(x, y) = -3 x + y - 1 and Q(x, y) = x + y + 3.

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(-3 x + y - 1) dx = -x - (3 x^2)/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 - (3 x^2)/2 + y x + g(y)) = x + (dg(y))/(dy)

Substitute into (df(x, y))/(dy) = Q(x, y):

x + (dg(y))/(dy) = x + y + 3

Solve for (dg(y))/(dy):

(dg(y))/(dy) = y + 3

Integrate (dg(y))/(dy) with respect to y:

g(y) = integral(y + 3) dy = y^2/2 + 3 y

Substitute g(y) into f(x, y):

f(x, y) = -(3 x^2)/2 - x + y^2/2 + 3 y + y x

The solution is f(x, y) = c_1:

-(3 x^2)/2 - x + y^2/2 + 3 y + y x = c_1

Solve for y:

y(x) = -x - sqrt(4 x^2 + 8 x + 2 c_1 + 9) - 3 or y(x) = -x + sqrt(4 x^2 + 8 x + 2 c_1 + 9) - 3

Simplify the arbitrary constants:

Answer: |

| y(x) = -x - sqrt(4 x^2 + 8 x + c_1) - 3 or y(x) = -x + sqrt(4 x^2 + 8 x + c_1) - 3