Solve the given boundary-value problem. (If an answer does not exist, enter DNE.) $$ y^{\prime \prime}-2 y^{\prime}+2 y=0, y(0)=1, y(\pi)=1 $$ $$ y(x)=\square $$

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UpStudy AI · Accepted Answer

To solve the boundary-value problem:

$$
y'' - 2y' + 2y = 0, \quad y(0) = 1, \quad y(\pi) = 1
$$

**Step 1: Find the General Solution**

First, solve the homogeneous differential equation:

$$
y'' - 2y' + 2y = 0
$$

The characteristic equation is:

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

Solving for $$ r $$:

$$
r = \frac{2 \pm \sqrt{(-4)}}{2} = 1 \pm i
$$

Thus, the general solution is:

$$
y(x) = e^x \left( A \cos x + B \sin x \right)
$$

**Step 2: Apply the Boundary Conditions**

1. **At $$ x = 0 $$:**

$$
y(0) = e^0 (A \cos 0 + B \sin 0) = A = 1
$$

So, $$ A = 1 $$.

2. **At $$ x = \pi $$:**

$$
y(\pi) = e^{\pi} \left( A \cos \pi + B \sin \pi \right) = e^{\pi} (-A + 0) = -A e^{\pi}
$$

Substituting $$ A = 1 $$:

$$
y(\pi) = -e^{\pi}
$$

However, the boundary condition requires $$ y(\pi) = 1 $$, which implies:

$$
-e^{\pi} = 1
$$

This is impossible since $$ e^{\pi} > 0 $$. Therefore, there is no solution that satisfies both boundary conditions.

**Conclusion:**

$$
y(x) = \text{DNE}
$$

**Answer:** DNE
There is no solution that satisfies both boundary conditions, so $$ y(x) = \text{DNE} $$.

Solve the given boundary-value problem. (If an answer does not exist, enter DNE.)

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