Differential equation

Question

Is the Different equation is an exact differential equation? İf it is not exact, what should be the real numbers “a and b” for this equation to have an integral factor of ?

In this case, find the general solution of the equation.

Answer 1

Collecting like terms we get: -3ydx - 6xdy = 0;

Now ∂/dy(-3y) = - 3 ≠ ∂/dx(-6x) = - 6, so thios is not exact equation.

Now we looking for integral factor μ(x,y) = x^ay^b, Then we get equation:

-3x^ay^b+1_{- 6x}^a+1y^b = 0.

To get exact equation we have now:∂/∂y(-3x^ay^b+1) = -3(b+1)x^ay^b = ∂/∂x(-6x^a+1y^b) = - 6(a+1)x^ay^b

b + 1 = 2(a + 1) or b = 2a + 1. If a = 1 then b = 3 and μ(x,y) = xy³ is integral factor

If a = 2 then b = 5 and μ(x,y) = x²y also integral factor ...;

Reducing by - 3 and multiplaying by xy³ we get xy⁴dx + 2x²y³dy = 0

∂F/∂x = xy⁴; F(x,y) = ∫xy⁴dx = x²y⁴/2 + φ(y);

Now ∂F/dy = 2x²y³ + φ'(y) = 2x²y³; φ'(y) = 0, φ(y) = C

So general solution is F(x,y) = x²y⁴/2 + C