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Please solve the following differential equation

\( \left[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right] \cos \

User MLEN
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Explanation:

To solve the given differential equation, let's denote the differential operator as \(L\):

\[L = \left[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right].\]

We want to find the solution to the equation \(L \cos \theta = 0\). To do this, we'll separate variables by assuming the solution is of the form \(\cos \theta = f(\theta) g(\phi)\), where \(f(\theta)\) depends only on \(\theta\) and \(g(\phi)\) depends only on \(\phi\).

Let's start by applying the operator \(L\) to \(\cos \theta\):

\[L \cos \theta = \left[\frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{\sin ^{2} \theta} \frac{\partial^{2}}{\partial \phi^{2}}\right] \cos \theta.\]

Using the product rule for differentiation, we can expand this expression:

\[L \cos \theta = \frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial \cos \theta}{\partial \theta}\right) + \frac{1}{\sin^2 \theta} \frac{\partial^2 \cos \theta}{\partial \phi^2}.\]

Now, let's compute the derivatives:

\[\frac{\partial \cos \theta}{\partial \theta} = -\sin \theta,\]

\[\frac{\partial^2 \cos \theta}{\partial \theta^2} = -\cos \theta,\]

\[\frac{\partial^2 \cos \theta}{\partial \phi^2} = 0.\]

Substituting these derivatives back into the expression, we have:

\[L \cos \theta = \frac{1}{\sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta (-\sin \theta)\right) + \frac{1}{\sin^2 \theta} \cdot 0.\]

Simplifying further:

\[L \cos \theta = -\frac{\partial}{\partial \theta}(\sin^2 \theta) = -2\sin \theta \cos \theta.\]

Now, we set \(L \cos \theta\) equal to zero:

\[-2\sin \theta \cos \theta = 0.\]

This equation is satisfied when \(\sin \theta = 0\) or \(\cos \theta = 0\). However, we're interested in the non-trivial solutions, so we disregard \(\sin \theta = 0\) (which corresponds to the trivial solution \(\cos \theta = 0\)).

Therefore, the solution to the given differential equation is \(\cos \theta = 0\) for any value of \(\theta\) except the trivial solution.

User Neurodynamic
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