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To verify that the piecewise-defined function \( y(x) \) is a solution to the differential equation \( xy' - 2y = 0 \), we need to perform the following steps:

### Given:
\[
y(x) = \begin{cases} 
-x^2 & \text{if } x < 0 \\
x^2 & \text{if } x \geq 0 
\end{cases}
\]
We need to show this satisfies the differential equation:
\[
xy' - 2y = 0
\]
for all \( x \in (-\infty, \infty) \).

### Step 1: Compute \( y'(x) \) (the derivative of \( y(x) \))

For \( x < 0 \):
\[
y(x) = -x^2 \implies y'(x) = \frac{d}{dx}(-x^2) = -2x
\]

For \( x \geq 0 \):
\[
y(x) = x^2 \implies y'(x) = \frac{d}{dx}(x^2) = 2x
\]

So, the derivative is:
\[
y'(x) = \begin{cases} 
-2x & \text{if } x < 0 \\
2x & \text{if } x \geq 0 
\end{cases}
\]

### Step 2: Plug into the differential equation

The differential equation is \( xy' - 2y = 0 \). Let's substitute \( y(x) \) and \( y'(x) \) for each case.

#### Case 1: \( x < 0 \)
For \( x < 0 \):
\[
y(x) = -x^2, \quad y'(x) = -2x
\]
Substituting into the differential equation:
\[
x(-2x) - 2(-x^2) = 0
\]
Simplifying:
\[
-2x^2 + 2x^2 = 0
\]
This holds true, so the equation is satisfied for \( x < 0 \).

#### Case 2: \( x \geq 0 \)
For \( x \geq 0 \):
\[
y(x) = x^2, \quad y'(x) = 2x
\]
Substituting into the differential equation:
\[
x(2x) - 2(x^2) = 0
\]
Simplifying:
\[
2x^2 - 2x^2 = 0
\]
This also holds true, so the equation is satisfied for \( x \geq 0 \).

### Conclusion:
The function \( y(x) \) satisfies the differential equation \( xy' - 2y = 0 \) for both \( x < 0 \) and \( x \geq 0 \). Therefore, \( y(x) \) is a solution to the given differential equation for all \( x \in (-\infty, \infty) \).

Would you like more details or have any questions?

Here are five related questions to expand on this topic:
1. What other types of piecewise functions can satisfy differential equations?
2. How does continuity of a piecewise function affect its differentiability?
3. Can a non-continuous function still solve a differential equation?
4. How would the solution change if the differential equation were \( xy' - 3y = 0 \)?
5. How do initial conditions influence the form of solutions for differential equations?

**Tip**: Always check the differentiability of a piecewise function at the boundaries where the pieces change. This can affect whether it solves the given equation across the entire domain.

Verify that the piecewise-defined function 
y(x) = { -x^2 for x < 0, x^2 for x ≥ 0 }
is a solution of the differential equation xy' - 2y = 0 where x ∈ (-∞, ∞).

Explore how to verify that a piecewise-defined function is a solution to a differential equation. Learn the step-by-step process of differentiating each piece of the function and substituting into the given differential equation. Suitable for undergraduate students studying Calculus.

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