Understanding the Problem

We are given an equation relating \(\theta\) and \(\varphi\), where \(\varphi\) is a function of \(\theta\). The equation is \(e^{\theta\varphi} = c + 4\theta\varphi\), where \(c\) is an arbitrary constant. We need to find an expression for \(\varphi \, d\theta\). This involves differentiating the given implicit relation.

Differentiating the Equation

We differentiate both sides of the given equation with respect to \(\theta\). Remember to use the chain rule and the product rule where necessary.

The equation is:

\(\qquad e^{\theta\varphi} = c + 4\theta\varphi\)

Differentiating both sides with respect to \(\theta\):

\(\qquad \frac{d}{d\theta}(e^{\theta\varphi}) = \frac{d}{d\theta}(c + 4\theta\varphi)\)

For the left side, we use the chain rule. The derivative of \(e^u\) with respect to \(\theta\) is \(e^u \frac{du}{d\theta}\). Here, \(u = \theta\varphi\). Using the product rule for \(\frac{d}{d\theta}(\theta\varphi)\), we get:

\(\qquad \frac{d}{d\theta}(\theta\varphi) = \frac{d\theta}{d\theta}\varphi + \theta\frac{d\varphi}{d\theta} = 1 \cdot \varphi + \theta\frac{d\varphi}{d\theta} = \varphi + \theta\frac{d\varphi}{d\theta}\)

So, the derivative of the left side is:

\(\qquad \frac{d}{d\theta}(e^{\theta\varphi}) = e^{\theta\varphi}\left(\varphi + \theta\frac{d\varphi}{d\theta}\right)\)

For the right side, the derivative of a constant \(c\) is 0. For the term \(4\theta\varphi\), we use the constant multiple rule and the product rule again:

\(\qquad \frac{d}{d\theta}(c + 4\theta\varphi) = \frac{d}{d\theta}(c) + 4\frac{d}{d\theta}(\theta\varphi) = 0 + 4\left(\varphi + \theta\frac{d\varphi}{d\theta}\right)\)

Equating the derivatives of both sides:

\(\qquad e^{\theta\varphi}\left(\varphi + \theta\frac{d\varphi}{d\theta}\right) = 4\left(\varphi + \theta\frac{d\varphi}{d\theta}\right)\)

Rearranging and Solving

Now, we rearrange the equation to find a relation between \(\varphi\) and \(\frac{d\varphi}{d\theta}\):

\(\qquad e^{\theta\varphi}\left(\varphi + \theta\frac{d\varphi}{d\theta}\right) - 4\left(\varphi + \theta\frac{d\varphi}{d\theta}\right) = 0\)

Factor out the common term \(\left(\varphi + \theta\frac{d\varphi}{d\theta}\right)\):

\(\qquad \left(e^{\theta\varphi} - 4\right)\left(\varphi + \theta\frac{d\varphi}{d\theta}\right) = 0\)

This equation holds true if either of the factors is zero:

• Case 1: \(e^{\theta\varphi} - 4 = 0 \implies e^{\theta\varphi} = 4\). Substituting this back into the original equation \(e^{\theta\varphi} = c + 4\theta\varphi\), we get \(4 = c + 4\theta\varphi\). This implies \(4\theta\varphi = 4-c\), which means \(\theta\varphi\) is a constant. If \(\theta\varphi = K\) (where \(K\) is a constant), differentiating with respect to \(\theta\) gives \(\frac{d}{d\theta}(\theta\varphi) = \frac{dK}{d\theta} \implies \varphi + \theta\frac{d\varphi}{d\theta} = 0\).
• Case 2: \(\varphi + \theta\frac{d\varphi}{d\theta} = 0\).

Both cases lead to the same differential relation:

\(\qquad \varphi + \theta\frac{d\varphi}{d\theta} = 0\)

We need to express \(\varphi \, d\theta\). From the relation above, we can write:

\(\qquad \varphi = -\theta\frac{d\varphi}{d\theta}\)

Multiplying both sides by \(d\theta\) (or considering the differential form directly from \(\varphi + \theta\frac{d\varphi}{d\theta} = 0\), which is \(\varphi \, d\theta + \theta \, d\varphi = 0\)), we get:

\(\qquad \varphi \, d\theta = -\theta \, d\varphi\)

Thus, \(\varphi \, d\theta\) is equal to \(-\theta \, d\varphi\). This is the resulting expression derived from the given differential equation.