Question

(2x+3y)dx+(3x-2y)dy=0

Answer 1

(3x_2y)dy= -(2x+3y)dx

dy/dx = -(2x+3y)/(3x_2y)
integrating both sides thn solve it

Answer 2

Given:

• A differential equation  $(2x+3y)dx+(3x-2y)dy=0$  is given.

To Find:

• Solve the given differential equation.

Solution:

$(2x+3y)dx+(3x-2y)dy=0$

           $\frac{dy}{dx}=-\frac{2x+3y}{3x+2y}\\ \frac{dy}{dx} = - \frac{3}{2}+\frac{5}{2}\frac{x}{3x+2y}$

let’s substitute  $z=3x+2y$

Its derivative is

               $\frac{dz}{dx}=3+2\frac{dy}{dx}$

              $\frac{dy}{dx}=\frac{1}{2}\frac{dz}{dx}-\frac{3}{2}$

              $\frac{1}{2}\frac{dz}{dx}-\frac{3}{2}=-\frac{3}{2}+\frac{5}{2}\frac{x}{z}$

               $\frac{dz}{dx}=5\frac{x}{z}$

Now, This is separable

           $zdz=5xdx$

Integrating both sides, we get

           $\frac{1}{2}z^2=\frac{5}{2}x^2+c$

Re substitute

             $(3x+2y)^2=5x^2+c$

              $9x^2+4y^2+12xy=5x^2+c$

              $4x^2+4y^2+12xy=c$

Thus,

            $4x^2+4y^2+12xy=c$