Understanding the Differential Equation

The given equation is \({e^{y\sqrt {1 - {x^2}} + x\sqrt {1 - {y^2}} }} = c{e^x}\). To find the degree and order of the differential equation that satisfies this relation, we first need to simplify the given equation and then differentiate it.

Simplifying the Equation

We can simplify the equation by taking the natural logarithm on both sides:

\(\ln\left({e^{y\sqrt {1 - {x^2}} + x\sqrt {1 - {y^2}} }}\right) = \ln\left(c{e^x}\right)\)

Using logarithm properties, \(\ln(e^A) = A\) and \(\ln(AB) = \ln A + \ln B\):

\(y\sqrt {1 - {x^2}} + x\sqrt {1 - {y^2}} = \ln c + \ln(e^x)\)

\(y\sqrt {1 - {x^2}} + x\sqrt {1 - {y^2}} = \ln c + x\)

Let \(K = \ln c\). Since \(c > 0\), \(K\) is a constant.

The simplified equation is:

\(y\sqrt {1 - {x^2}} + x\sqrt {1 - {y^2}} = K + x\)

This is an implicit relation between \(x\) and \(y\). We need to differentiate this with respect to \(x\) to find the differential equation.

Differentiating the Implicit Equation

We differentiate the simplified equation \(y\sqrt {1 - {x^2}} + x\sqrt {1 - {y^2}} = K + x\) with respect to \(x\).

Using the product rule \((uv)' = u'v + uv'\) and the chain rule:

\(\frac{d}{dx}(y\sqrt {1 - {x^2}}) + \frac{d}{dx}(x\sqrt {1 - {y^2}}) = \frac{d}{dx}(K + x)\)

For the first term, \(u=y\), \(v=\sqrt{1-x^2}=(1-x^2)^{1/2}\):

\(\frac{dy}{dx}\sqrt{1-x^2} + y \cdot \frac{1}{2}(1-x^2)^{-1/2}(-2x) = \frac{dy}{dx}\sqrt{1-x^2} - \frac{xy}{\sqrt{1-x^2}}\)

For the second term, \(u=x\), \(v=\sqrt{1-y^2}=(1-y^2)^{1/2}\). Remember that \(y\) is a function of \(x\), so we use the chain rule for \(v'\):

\(1 \cdot \sqrt{1-y^2} + x \cdot \frac{1}{2}(1-y^2)^{-1/2}(-2y)\frac{dy}{dx} = \sqrt{1-y^2} - \frac{xy}{\sqrt{1-y^2}}\frac{dy}{dx}\)

The derivative of the right side \(K+x\) is \(0 + 1 = 1\).

Combining the terms, the differential equation is:

\(\frac{dy}{dx}\sqrt{1-x^2} - \frac{xy}{\sqrt{1-x^2}} + \sqrt{1-y^2} - \frac{xy}{\sqrt{1-y^2}}\frac{dy}{dx} = 1\)

Determining the Order and Degree

To determine the order and degree, we need to identify the highest derivative present and its highest power after writing the equation as a polynomial in the derivatives.

In the equation \(\frac{dy}{dx}\sqrt{1-x^2} - \frac{xy}{\sqrt{1-x^2}} + \sqrt{1-y^2} - \frac{xy}{\sqrt{1-y^2}}\frac{dy}{dx} = 1\), the highest derivative present is \(\frac{dy}{dx}\), which is a first derivative.

Therefore, the order of the differential equation is 1.

Now, let's examine the degree. We rearrange the equation to group the terms involving \(\frac{dy}{dx}\):

\(\frac{dy}{dx}\left(\sqrt{1-x^2} - \frac{xy}{\sqrt{1-y^2}}\right) = 1 + \frac{xy}{\sqrt{1-x^2}} - \sqrt{1-y^2}\)

Let \(p = \frac{dy}{dx}\). The equation is \(p\left(\sqrt{1-x^2} - \frac{xy}{\sqrt{1-y^2}}\right) = 1 + \frac{xy}{\sqrt{1-x^2}} - \sqrt{1-y^2}\).

This equation is a polynomial in \(p\), and the highest power of \(p\) is 1. The terms multiplying \(p\) and the terms on the right side involve \(x\), \(y\), \(\sqrt{1-x^2}\), and \(\sqrt{1-y^2}\), but they do not involve the derivative \(p\).

To be formal about the degree, we can clear the denominators involving the square roots:

\(p\left(\frac{\sqrt{1-x^2}\sqrt{1-y^2} - xy}{\sqrt{1-y^2}}\right) = \frac{\sqrt{1-x^2} + xy - \sqrt{1-x^2}\sqrt{1-y^2}}{\sqrt{1-x^2}}\)

Multiplying both sides by \(\sqrt{1-x^2}\sqrt{1-y^2}\):

\(p (\sqrt{1-x^2}\sqrt{1-y^2} - xy)\sqrt{1-x^2} = (\sqrt{1-x^2} + xy - \sqrt{1-x^2}\sqrt{1-y^2})\sqrt{1-y^2}\)

This equation is a polynomial in \(p\). The highest power of \(p\) is 1.

Therefore, the degree of the differential equation is 1.

The degree and order are 1 and 1, respectively.