1. **Problem Statement:** Obtain the differential equation from the relation $y = cx + c^2 + 1$. Show that $y_1 = e^x$ and $y_2 = xe^x$ form a fundamental set of solutions of the differential equation $$y'' - 2y' + y = 0.$$ 2. **Finding the differential equation from $y = cx + c^2 + 1$: ** - Given $y = cx + c^2 + 1$, where $c$ is an arbitrary constant. - Differentiate $y$ with respect to $x$: $$y' = c$$ - Differentiate again: $$y'' = 0$$ - From $y' = c$, we get $c = y'$. Substitute into original equation: $$y = y' x + (y')^2 + 1$$ - Rearranged: $$y - x y' - (y')^2 = 1$$ - This is the implicit differential equation relating $x$, $y$, and $y'$. 3. **Showing $y_1 = e^x$ and $y_2 = x e^x$ form a fundamental set of solutions of $$y'' - 2y' + y = 0$$:** - Compute derivatives for $y_1 = e^x$: $$y_1' = e^x, \quad y_1'' = e^x$$ - Substitute into the differential equation: $$y_1'' - 2 y_1' + y_1 = e^x - 2 e^x + e^x = 0$$ - Compute derivatives for $y_2 = x e^x$: $$y_2' = e^x + x e^x = (1 + x) e^x$$ $$y_2'' = e^x + e^x + x e^x = (2 + x) e^x$$ - Substitute into the differential equation: $$y_2'' - 2 y_2' + y_2 = (2 + x) e^x - 2 (1 + x) e^x + x e^x = (2 + x - 2 - 2x + x) e^x = 0$$ - Since both satisfy the equation, they are solutions. 4. **Check linear independence (Wronskian):** - Wronskian $W = y_1 y_2' - y_2 y_1' = e^x (1 + x) e^x - x e^x e^x = e^{2x} (1 + x - x) = e^{2x} \neq 0$ - Since $W \neq 0$, $y_1$ and $y_2$ form a fundamental set of solutions. **Final answers:** - Differential equation from $y = cx + c^2 + 1$ is $$y - x y' - (y')^2 = 1.$$ - $y_1 = e^x$ and $y_2 = x e^x$ form a fundamental set of solutions of $$y'' - 2 y' + y = 0.$$