The differential equation 2y dx – (3y – 2x) dy = 0 is

Let's analyze the given differential equation to determine if it is exact, homogeneous, and linear.

Differential Equation Analysis

The given differential equation is:

$$2ydx-(3y-2x)dy=0$$

We can rewrite this equation in the standard form $M(x,y)dx+N(x,y)dy=0$.

First, distribute the negative sign:

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

Rearranging the terms in $N(x,y)$:

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

From this, we identify:

- $M(x,y)=2y$
- $N(x,y)=2x-3y$

Determining if the Equation is Homogeneous

A differential equation $M(x,y)dx+N(x,y)dy=0$ is called homogeneous if both $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree. A function $f(x,y)$ is homogeneous of degree $n$ if $f(kx,ky)=k^{n}f(x,y)$ for some constant $n$.

Let's check $M(x,y)$:

$$M(kx,ky)=2(ky)=k(2y)=k^{1}M(x,y)$$

So, $M(x,y)$ is homogeneous of degree 1.

Now let's check $N(x,y)$:

$$N(kx,ky)=2(kx)-3(ky)=k(2x-3y)=k^{1}N(x,y)$$

So, $N(x,y)$ is also homogeneous of degree 1.

Since both $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree (1), the differential equation is homogeneous.

Determining if the Equation is Exact

A differential equation $M(x,y)dx+N(x,y)dy=0$ is exact if the partial derivative of $M$ with respect to $y$ equals the partial derivative of $N$ with respect to $x$. That is, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.

Calculate the partial derivatives:

- $$\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}(2y)=2$$
- $$\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}(2x-3y)=2$$

Since $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ (both equal 2), the differential equation is exact.

Determining if the Equation is Linear

A first-order differential equation is linear if it can be expressed in the form $\frac{dy}{dx}+P(x)y=Q(x)$ or $\frac{dx}{dy}+P(y)x=Q(y)$.

Let's rearrange the given equation to see if it fits the linear form. It's often easier to check by writing it as $\frac{dx}{dy}$:

$$2ydx=(3y-2x)dy$$

Divide both sides by $2ydy$ (assuming $y\ne 0$):

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

Separate the terms on the right side:

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

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

Rearrange to match the standard linear form $\frac{dx}{dy}+P(y)x=Q(y)$:

$$\frac{dx}{dy}+\frac{1}{y}x=\frac{3}{2}$$

This equation is in the standard form of a linear differential equation, where $P(y)=\frac{1}{y}$ and $Q(y)=\frac{3}{2}$. Therefore, the differential equation is linear (specifically, linear in $x$ as a function of $y$).

Conclusion

Based on the analysis, the differential equation $2ydx-(3y-2x)dy=0$ satisfies all three properties:

- It is exact because $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$.
- It is homogeneous because $M(x,y)$ and $N(x,y)$ are homogeneous functions of the same degree (1).
- It is linear because it can be written in the standard linear form $\frac{dx}{dy}+P(y)x=Q(y)$.

Therefore, the correct classification is exact, homogeneous, and linear.