Two linearly independent solutions of the differential equation $y ^ { \prime \prime } - 4 y ^ { \prime } + 4 y = 0$ are
A) $y _ { 1 } = e ^ { 2 x } , y _ { 2 } = e ^ { 2 x }$
B) $y _ { 1 } = e ^ { 2 x } , y _ { 2 } = x e ^ { 2 x }$
C) $y _ { 1 } = e ^ { 2 x } , y _ { 2 } = e ^ { - 2 x }$
D) $y _ { 1 } = e ^ { - 2 x } , y _ { 2 } = x e ^ { - 2 x }$
E) $y _ { 1 } = e ^ { - 2 x } , y _ { 2 } = x e ^ { 2 x }$

Question

Quizplus · Accepted Answer

The characteristic equation of the given differential equation $y'' - 4y' + 4y = 0$ is $r^2 - 4r + 4 = 0$, which simplifies to $(r - 2)^2 = 0$. This equation has a repeated root $r = 2$. For differential equations with repeated roots, the solutions are of the form $y_1 = e^{rx}$ and $y_2 = xe^{rx}$. Therefore, the correct solutions are $y_1 = e^{2x}$ and $y_2 = xe^{2x}$.

Two Linearly Independent Solutions of the Differential Equation y′′−4y′+4y=0y ^ { \prime \prime } - 4 y ^ { \prime } + 4 y = 0y′′−4y′+4y=0

Two Linearly Independent Solutions of the Differential Equation $y ^ { \prime \prime } - 4 y ^ { \prime } + 4 y = 0$