Solve the differential equation du/dx + k*u*x^n =0, using a power series expansion u(x) = \sum_{m=0}^{\infty}{a_m x^m}, i.e. find the coefficients a_m.   This method sometimes known as the method of Frobenius. 

To solve the differential equation

\[
\frac{du}{dx} + k u x^n = 0
\]

using a power series expansion, \( u(x) = \sum_{m=0}^{\infty} a_m x^m \), we apply the method of Frobenius. Here's the step-by-step approach:

### Step 1: Express \( u(x) \) and its derivative as power series

Let \( u(x) = \sum_{m=0}^{\infty} a_m x^m \).

Now, differentiate \( u(x) \) term by term:

\[
\frac{du}{dx} = \sum_{m=1}^{\infty} m a_m x^{m-1}.
\]

### Step 2: Substitute into the differential equation

The original equation is:

\[
\frac{du}{dx} + k u x^n = 0.
\]

Substitute \( u(x) = \sum_{m=0}^{\infty} a_m x^m \) and \( \frac{du}{dx} = \sum_{m=1}^{\infty} m a_m x^{m-1} \) into this equation:

\[
\sum_{m=1}^{\infty} m a_m x^{m-1} + k \left( \sum_{m=0}^{\infty} a_m x^m \right) x^n = 0.
\]

### Step 3: Adjust the powers of \( x \)

Now, expand the second term:

\[
k \left( \sum_{m=0}^{\infty} a_m x^m \right) x^n = k \sum_{m=0}^{\infty} a_m x^{m+n}.
\]

This gives the full equation:

\[
\sum_{m=1}^{\infty} m a_m x^{m-1} + k \sum_{m=0}^{\infty} a_m x^{m+n} = 0.
\]

### Step 4: Align powers of \( x \)

We now need to match powers of \( x \). Rewrite the sums so that both sums have the same power of \( x \). In the first sum, let’s shift the index by setting \( m-1 = j \), or \( m = j+1 \):

\[
\sum_{j=0}^{\infty} (j+1) a_{j+1} x^j + k \sum_{m=0}^{\infty} a_m x^{m+n} = 0.
\]

This can be written as:

\[
\sum_{j=0}^{\infty} (j+1) a_{j+1} x^j + k \sum_{m=0}^{\infty} a_m x^{m+n} = 0.
\]

### Step 5: Combine the sums and solve for the coefficients

To simplify, we combine terms with the same power of \( x \). The equation becomes:

\[
\sum_{j=0}^{\infty} \left[ (j+1) a_{j+1} + k a_{j-n} \right] x^j = 0.
\]

For this sum to be zero for all \( x^j \), the coefficients must satisfy:

\[
(j+1) a_{j+1} + k a_{j-n} = 0.
\]

### Step 6: Find the recurrence relation for \( a_m \)

This gives a recurrence relation for the coefficients:

\[
a_{j+1} = -\frac{k}{j+1} a_{j-n}.
\]

Thus, we can express each \( a_{m} \) in terms of the previous coefficients, depending on \( k \) and \( n \).

### Summary of the Solution

The recurrence relation for the coefficients \( a_m \) is:

\[
a_{m+1} = -\frac{k}{m+1} a_{m-n}.
\]

This provides the general form for the coefficients, allowing you to recursively calculate \( a_m \) from initial conditions or boundary values.

Would you like further details or clarifications on specific steps?

### Related Questions:
1. How can you choose initial conditions for the series solution?
2. How does the value of \( n \) influence the solution?
3. Can you provide an example using specific values of \( k \) and \( n \)?
4. What happens if \( n = 0 \) (i.e., when the differential equation is linear)?
5. How would the series solution behave for large values of \( x \)?

### Tip:
When solving differential equations using power series, pay close attention to the recurrence relations as they define how each coefficient depends on the previous terms.

Solve the differential equation du/dx + k*u*x^n =0, using a power series expansion u(x) = \sum_{m=0}^{\infty}{a_m x^m}, i.e., find the coefficients a_m. This method is sometimes known as the method of Frobenius.

Learn how to solve the differential equation du/dx + k*u*x^n = 0 using a power series expansion and the method of Frobenius. This detailed step-by-step solution involves key concepts such as power series, recurrence relations, and the Frobenius method, providing a thorough understanding for undergraduate mathematics or engineering students.

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