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The given problem appears to be a second-order linear differential equation of the form:

\[
2y'' + y = 0
\]

Along with the initial conditions:

\[
y = 1, \quad y' = \sqrt{2} \quad \text{at} \quad x = \frac{\pi}{\sqrt{2}}.
\]

### Step 1: Solve the Homogeneous Differential Equation

This is a second-order linear differential equation with constant coefficients. The general solution approach involves solving the characteristic equation.

For the equation:
\[
2y'' + y = 0
\]

Divide by 2 to simplify:
\[
y'' + \frac{1}{2}y = 0
\]

Now, assume a solution of the form \(y = e^{rx}\). Substituting into the differential equation:

\[
r^2 e^{rx} + \frac{1}{2} e^{rx} = 0
\]

Factoring out \(e^{rx}\), which is never zero:
\[
r^2 + \frac{1}{2} = 0
\]

The characteristic equation is:
\[
r^2 = -\frac{1}{2}
\]

Thus, \(r = \pm i\frac{1}{\sqrt{2}}\).

### Step 2: General Solution

The solution to this characteristic equation involves sinusoidal terms:
\[
y(x) = c_1 \cos\left(\frac{x}{\sqrt{2}}\right) + c_2 \sin\left(\frac{x}{\sqrt{2}}\right)
\]

### Step 3: Apply Initial Conditions

We are given the following initial conditions:

\[
y\left(\frac{\pi}{\sqrt{2}}\right) = 1, \quad y'\left(\frac{\pi}{\sqrt{2}}\right) = \sqrt{2}
\]

#### Find \(y'\):

Differentiating \(y(x)\):
\[
y'(x) = -\frac{1}{\sqrt{2}} c_1 \sin\left(\frac{x}{\sqrt{2}}\right) + \frac{1}{\sqrt{2}} c_2 \cos\left(\frac{x}{\sqrt{2}}\right)
\]

#### Use \(y\left(\frac{\pi}{\sqrt{2}}\right) = 1\):

Substitute \(x = \frac{\pi}{\sqrt{2}}\) into the general solution:
\[
1 = c_1 \cos\left(\frac{\pi}{\sqrt{2} \cdot \sqrt{2}}\right) + c_2 \sin\left(\frac{\pi}{\sqrt{2} \cdot \sqrt{2}}\right)
\]
This simplifies to:
\[
1 = c_1 \cos\left(\frac{\pi}{2}\right) + c_2 \sin\left(\frac{\pi}{2}\right)
\]
\[
1 = c_2
\]

#### Use \(y'\left(\frac{\pi}{\sqrt{2}}\right) = \sqrt{2}\):

Substitute into \(y'(x)\):
\[
\sqrt{2} = -\frac{1}{\sqrt{2}} c_1 \sin\left(\frac{\pi}{2}\right) + \frac{1}{\sqrt{2}} c_2 \cos\left(\frac{\pi}{2}\right)
\]
\[
\sqrt{2} = -\frac{1}{\sqrt{2}} c_1 \cdot 1 + 0
\]
\[
\sqrt{2} = -\frac{1}{\sqrt{2}} c_1
\]
\[
c_1 = -2
\]

### Step 4: Final Solution

Substitute the values of \(c_1\) and \(c_2\) into the general solution:

\[
y(x) = -2 \cos\left(\frac{x}{\sqrt{2}}\right) + \sin\left(\frac{x}{\sqrt{2}}\right)
\]

This is the solution to the given differential equation with the specified initial conditions.

---

Would you like further clarification or details on any of the steps?

Here are five related questions you could explore next:

1. What is the general approach to solving second-order differential equations with constant coefficients?
2. How do complex roots of characteristic equations affect the form of the solution?
3. How can initial conditions be applied to systems of differential equations?
4. What are some physical systems that can be modeled by second-order differential equations?
5. How does the method of undetermined coefficients differ from the characteristic equation method?

**Tip:** When solving differential equations, always verify that your final solution satisfies both the differential equation and the initial/boundary conditions.

Solve the second-order differential equation 2y'' + y = 0 with the initial conditions y = 1 and y' = √2 at x = π/√2.

Solve the second-order differential equation 2y'' + y = 0 with initial conditions y = 1 and y' = √2 at x = π/√2. Learn how to apply the characteristic equation method and solve for constants using initial conditions.

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