Find the general solution of y'' - 2y' + y = 0
Find the general solution of $y'' - 2y' + y = 0$.
Explanation of Method
A repeated real root of the characteristic equation forces a second solution built by multiplying the first by x.
Step 1: Characteristic equation
Substituting $y=e^{rx}$: $r^2-2r+1=0.$
The left side is a perfect square, $(r-1)^2=r^2-2r+1$, confirmed by expansion, so: \[(r-1)^2=0 \implies r_1=r_2=1,\]
a single real root of multiplicity two.
Step 2: Fundamental system
A repeated real root $r=1$ contributes one solution $e^{x}$ outright; the second independent solution is obtained by multiplying by $x$, giving
\[ y_1=e^{x}, \qquad y_2=xe^{x}. \]
Step 3: Wronskian
\[ W(y_1,y_2)=\begin{vmatrix}e^x & xe^x\\e^x & e^x+xe^x\end{vmatrix} = e^x(e^x+xe^x)-xe^x\cdot e^x = e^{2x} \neq 0, \]
confirming $y_1,y_2$ are linearly independent for all $x$.
Step 4: General solution
The solution is recorded in the result below.
Result
Final answer:
\[\boxed{\,y = C_1e^{x} + C_2xe^{x}\,}\]