6. Find the general solution of $$ y^{\prime \prime}-2 y^{\prime}+y=0 $$ $$ \begin{array}{l}\text { A. } y(x)=c_{1} e^{x}+c_{2} x e^{x} \ \text { B. } y(x)=c_{1} e^{x}+c_{2} e^{x} \ \text { C. } y(x)=c_{1} x+c_{2} x \ln x \ \text { D. } y(x)=c_{1} x+c_{4} x^{-1}\end{array} $$

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UpStudy AI · Accepted Answer

To find the general solution of the differential equation

$$
y^{\prime \prime} - 2y^{\prime} + y = 0,
$$

we start by finding the characteristic equation associated with this linear homogeneous differential equation. The characteristic equation is obtained by substituting $$ y = e^{rx} $$ into the differential equation, leading to:

$$
r^2 - 2r + 1 = 0.
$$

This can be factored as:

$$
(r - 1)^2 = 0.
$$

This indicates that we have a repeated root $$ r = 1 $$. For a second-order linear differential equation with constant coefficients, if we have a repeated root $$ r $$, the general solution takes the form:

$$
y(x) = (c_1 + c_2 x)e^{rx}.
$$

Substituting $$ r = 1 $$, we get:

$$
y(x) = (c_1 + c_2 x)e^{x}.
$$

This can be rewritten as:

$$
y(x) = c_1 e^{x} + c_2 x e^{x}.
$$

Now, we can compare this with the provided options:

- A. $$ y(x) = c_1 e^{x} + c_2 x e^{x} $$
- B. $$ y(x) = c_1 e^{x} + c_2 e^{x} $$
- C. $$ y(x) = c_1 x + c_2 x \ln x $$
- D. $$ y(x) = c_1 x + c_4 x^{-1} $$

The correct answer is:

$$
\boxed{A}
$$

$$ y(x) = c_1 e^{x} + c_2 x e^{x} $$.
The general solution is $$ y(x) = c_1 e^{x} + c_2 x e^{x} $$. Therefore, the correct answer is A.

6. Find the general solution of \( y^{\prime \prime}-2 y^{\prime}+y=0 \) \[ \begin{array}{l}\text { A. } y(x)=c_{1} e^{x}+c_{2} x e^{x} \\ \text { B. } y(x)=c_{1} e^{x}+c_{2} e^{x} \\ \text { C. } y(x)=c_{1} x+c_{2} x \ln x \\ \text { D. } y(x)=c_{1} x+c_{4} x^{-1}\end{array} \]

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