Differential Equation Classification: Detailed Analysis

To classify the given differential equation, \(\left( {{y^2}{e^{x{y^2}}} + 6x} \right)dx + \left( {2xy{e^{x{y^2}}} - 4y} \right)dy = 0\), we need to determine if it is linear or non-linear, homogeneous or non-homogeneous, and exact or inexact. This analysis involves examining the form of the equation and applying specific tests for each property.

Linearity of the Differential Equation

A first-order differential equation is considered linear if it can be written in the form \(\frac{{dy}}{{dx}} + P(x)y = Q(x)\) or \(\frac{{dx}}{{dy}} + P(y)x = Q(y)\). Key characteristics of a linear equation include:

• The dependent variable and its derivatives appear only to the first power.
• There are no products of the dependent variable and its derivatives.
• There are no transcendental functions (like \(e^y\), \(\sin(y)\), etc.) of the dependent variable.

The given equation is \(\left( {{y^2}{e^{x{y^2}}} + 6x} \right)dx + \left( {2xy{e^{x{y^2}}} - 4y} \right)dy = 0\). Let's examine the terms:

• The term \(y^2e^{xy^2}\) contains \(y^2\) and \(xy^2\) in the exponent. The presence of \(y^2\) and the product \(xy^2\) makes the equation non-linear with respect to \(y\).
• Similarly, the term \(2xye^{xy^2}\) contains the product \(xy\) and \(xy^2\) in the exponent, further confirming its non-linear nature.

Since the dependent variable \(y\) appears in terms like \(y^2\) and as an argument of an exponential function with products involving \(y\) (i.e., \(xy^2\)), the differential equation is non-linear.

Homogeneity of the Differential Equation

A differential equation of the form \(M(x,y)dx + N(x,y)dy = 0\) is 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 \(f(tx, ty) = t^n f(x,y)\) for some constant \(n\).

In our given equation:

• \(M(x,y) = y^2 e^{xy^2} + 6x\)
• \(N(x,y) = 2xy e^{xy^2} - 4y\)

Let's test \(M(x,y)\) for homogeneity:

Substitute \(x \rightarrow tx\) and \(y \rightarrow ty\):

\(M(tx, ty) = (ty)^2 e^{(tx)(ty)^2} + 6(tx)\)

\(M(tx, ty) = t^2 y^2 e^{t^3 xy^2} + 6tx\)

For \(M(x,y)\) to be homogeneous, \(M(tx, ty)\) should be equal to \(t^n M(x,y)\). However, due to the presence of \(t^3\) inside the exponent of \(e^{t^3 xy^2}\), this expression cannot be factored into \(t^n M(x,y)\). Specifically, the exponential function \(e^{xy^2}\) is not a homogeneous function.

Therefore, the differential equation is non-homogeneous.

Exactness of the Differential Equation

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 \(x\). This is the condition: \(\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}\).

Given:

• \(M(x,y) = y^2 e^{xy^2} + 6x\)
• \(N(x,y) = 2xy e^{xy^2} - 4y\)

Let's calculate \(\frac{{\partial M}}{{\partial y}}\):

\(\frac{{\partial M}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {y^2 e^{xy^2} + 6x} \right)\)

Using the product rule \(\frac{d}{dy}(uv) = u'v + uv'\) for \(y^2 e^{xy^2}\):

• Let \(u = y^2 \Rightarrow u' = 2y\)
• Let \(v = e^{xy^2} \Rightarrow v' = e^{xy^2} \cdot \frac{\partial}{\partial y}(xy^2) = e^{xy^2} \cdot (2xy)\)

So,

\(\frac{{\partial M}}{{\partial y}} = (2y)e^{xy^2} + y^2(2xy e^{xy^2}) + 0\)

\(\frac{{\partial M}}{{\partial y}} = 2y e^{xy^2} + 2xy^3 e^{xy^2}\)

\(\frac{{\partial M}}{{\partial y}} = e^{xy^2}(2y + 2xy^3)\)

Now, let's calculate \(\frac{{\partial N}}{{\partial x}}\):

\(\frac{{\partial N}}{{\partial x}} = \frac{\partial }{{\partial x}}\left( {2xy e^{xy^2} - 4y} \right)\)

Using the product rule \(\frac{d}{dx}(uv) = u'v + uv'\) for \(2xy e^{xy^2}\):

• Let \(u = 2xy \Rightarrow u' = 2y\) (treating \(y\) as a constant for partial differentiation with respect to \(x\))
• Let \(v = e^{xy^2} \Rightarrow v' = e^{xy^2} \cdot \frac{\partial}{\partial x}(xy^2) = e^{xy^2} \cdot (y^2)\)

So,

\(\frac{{\partial N}}{{\partial x}} = (2y)e^{xy^2} + (2xy)(y^2 e^{xy^2}) - 0\)

\(\frac{{\partial N}}{{\partial x}} = 2y e^{xy^2} + 2xy^3 e^{xy^2}\)

\(\frac{{\partial N}}{{\partial x}} = e^{xy^2}(2y + 2xy^3)\)

Comparing the two partial derivatives:

\(\frac{{\partial M}}{{\partial y}} = e^{xy^2}(2y + 2xy^3)\)

\(\frac{{\partial N}}{{\partial x}} = e^{xy^2}(2y + 2xy^3)\)

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

Conclusion: Properties of the Differential Equation

Based on our detailed analysis, we can conclude the following about the given differential equation \(\left( {{y^2}{e^{x{y^2}}} + 6x} \right)dx + \left( {2xy{e^{x{y^2}}} - 4y} \right)dy = 0\):

• It is Non-linear due to the presence of terms like \(y^2\) and \(xy^2\) within an exponential function.
• It is Non-homogeneous because the functions \(M(x,y)\) and \(N(x,y)\) do not satisfy the homogeneity condition \(f(tx, ty) = t^n f(x,y)\).
• It is Exact because the condition \(\frac{{\partial M}}{{\partial y}} = \frac{{\partial N}}{{\partial x}}\) is satisfied.

Therefore, the differential equation is Non-linear; non-homogeneous and exact.