Question

​​​​​​​ (1 point) If \[ y=\sum_{n=0}^{\infty} c_{n} x^{n} \] is a solution of the differential equation \[ y^{\prime \prime}+(-x+1) y^{\prime}-3 y=0 \] then its coefficients \( c_{n} \) are related by the equ

Answer 1

The coefficients
`\( c_n \)` are related by the recurrence relation
`\( c_(n+2) = ((n+3)(n+2))/((n+2)(n+1) + 3) c_n \).`

Step-by-step explanation:

To find the relationship between the coefficients
`\( c_n \)`, let's substitute the power series
`\( y = \sum_(n=0)^(\infty) c_(n) x^(n) \)` into the given differential equation
`\( y^(\prime \prime)+(-x+1) y^(\prime)-3 y=0 \)` and solve for
`\( c_n \)`.

Firstly, calculate the derivatives:

`\[ y' = \sum_(n=0)^(\infty) n c_n x^(n-1) \]`

`\[ y'' = \sum_(n=0)^(\infty) n(n-1) c_n x^(n-2) \]`

Now, substitute these into the differential equation:

`\[ \sum_(n=0)^(\infty) n(n-1) c_n x^(n-2) + (-x+1)\sum_(n=0)^(\infty) n c_n x^(n-1) - 3\sum_(n=0)^(\infty) c_(n) x^(n) = 0 \]`

Combine the terms and regroup:

`\[ \sum_(n=0)^(\infty) (n(n-1) c_n + n c_n - 3c_n) x^(n-2) + \sum_(n=0)^(\infty) (-x + 1)n c_n x^(n-1) = 0 \]`

This yields the following recurrence relation:

`\[ n(n-1) c_n + n c_n - 3c_n = 0 \]`

Simplify this relation:

`\[ c_(n+2) = ((n+3)(n+2))/((n+2)(n+1) + 3) c_n \]`

Thus, the coefficients
`\( c_n \)` are related by the recurrence relation
`\( c_(n+2) = ((n+3)(n+2))/((n+2)(n+1) + 3) c_n \).`