Question

Classify the DE as separable, linear, exact, or none of these. An equation may have more than one classification. (x⁴/³ -3y) dx + xdy=0

a) separable
b) linear if y is the dependent variable
c) linear if x is the dependent variable
d) none of these
e) exact

Answer 1

The given differential equation cannot be classified as separable, linear, or exact.

Step-by-step explanation:

The given differential equation is (x4/3 - 3y)dx + xdy = 0.

To classify the DE, we need to check if it is separable, linear, exact, or none of these.

Let's check each classification:

- Separable: A separable DE can be expressed as P(x)dx + Q(y)dy = 0, where P(x) and Q(y) are functions of x and y, respectively. The given DE cannot be expressed in this form.

- Linear: A linear DE can be expressed as P(x)y' + Q(x)y = R(x), where P(x), Q(x), and R(x) are functions of x. The given DE does not satisfy this form.

- Exact: A DE is exact if it can be expressed in the form M(x, y)dx + N(x, y)dy = 0, where M(x, y) and N(x, y) are functions of both x and y, and ∂M/∂y = ∂N/∂x. To check if the given DE is exact, we need to calculate ∂M/∂y and ∂N/∂x and see if they are equal. Here, M(x, y) = (x4/3 - 3y) and N(x, y) = x. Differentiating, we have ∂M/∂y = -3 and ∂N/∂x = 1, which are not equal. Hence, the DE is not exact.

Based on the above analysis, the given DE can be classified as none of these.