1. **Problem statement:** Show that if $y = e^x$, then $$y_{n+1} - 2y_n - 2xy_n' + n(n - 1)y_{n-1} = 0.$$ 2. **Recall definitions and formulas:** Here, $y_n$ denotes the $n$th derivative of $y$ with respect to $x$, i.e., $y_n = \frac{d^n y}{dx^n}$. Since $y = e^x$, all derivatives are $e^x$, so $y_n = e^x$ for all $n$. Also, $y_n' = y_{n+1}$. 3. **Substitute known values:** Since $y_n = e^x$, $y_n' = y_{n+1} = e^x$, and $y_{n-1} = e^x$, substitute into the equation: $$y_{n+1} - 2y_n - 2xy_n' + n(n - 1)y_{n-1} = e^x - 2e^x - 2x e^x + n(n - 1) e^x.$$ 4. **Simplify:** Factor out $e^x$: $$e^x (1 - 2 - 2x + n(n - 1)) = e^x (n(n - 1) - 1 - 2x).$$ 5. **Analyze the expression:** For the equation to hold for all $x$, the expression inside parentheses must be zero: $$n(n - 1) - 1 - 2x = 0.$$ 6. **Conclusion:** This cannot be true for all $x$ unless $n(n - 1) - 1 = 2x$, which depends on $x$. Since $n$ is an integer and $x$ is variable, the original equation does not hold for all $x$ unless $x$ is fixed. **Therefore, the given equation is not generally true for $y = e^x$ unless additional context or constraints are provided.**