How do you categorize the following differential equation?

Question

Consider the following differential equation:

x * y' + y = 0

My task is to determine whether or not the following adjectives describe the equation above:

- Exact

- Homogeneous

- Heterogeneous

- Mixed Order

- Linear

- Autonomous

Thank you for any help you can provide!

Answer 1

Your differential equation is of the form:

a(x,y)*y' + b(x,y) = 0

Such a differential equation is "exact" iff:

da(x,y)/dx = db(x.y)/dy

where the derivatives in this equation are partial derivatives.

In this case, a(x,y) = x, so da/dx = 1, and b(x,y) = y, and db/dy = 1, so da/dx = db/dy = 1 and the equation is exact.

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This equation is a first-order equation (the highest order derivative of the independent variable appearing in the equation is dy/dx, which is a first derivative).  A first-order differential equation is homogeneous if it can be written in the form:

dy/dx = g(y/x)

where g is a function that has the quantity (y/x) as its independent variable.  Let's see if your equation can be put in this form.

x* y' + y = 0

x*y' = -y

y' = -(y/x)

If we let g(y/x) = -y/x, then it is clear that this equation is homogeneous.

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A differential equation that is not homogeneous is heterogeneous.  We just showed that this equation is homogeneous, so it can't be heterogeneous.

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I'm not sure what is meant by "mixed order" here.  I'm only familiar with this term when it's applied to systems of differential equations.  In any case, this is a first order equation, so it's almost certain that it's not "mixed order".

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A linear differential equation is one in which all the derivatives appear as linear factors (i.e., not raised to a power other than 1), and all the coefficients of the differential terms involve only the independent variable (or are constants).  In this case, y' and y are both linear factors, and their coefficients are x and 1, respectively, so this is a linear equation.

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An autonomous differential equation is one in which the independent variable does not appear explicitly.  For instance, a second-order autonomous equation is of the form F(y, y', y'') = 0, and an example would be y'' + 2y' - y^2 = 0.

The independent variable, x, appears explicitly (as the coefficient of the y' term) in the equation in this question, so it is not an autonomous equation.