To show that the equation $\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} - 2y = 0$ holds for $y = e^{x^2}$, let's follow these steps:

Step 1: Find the first derivative $\frac{dy}{dx}$

Given $y = e^{x^2}$, let's differentiate $y$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(e^{x^2})$

Using the chain rule, where the derivative of $e^{u}$ with respect to $u$ is $e^u$, and $u = x^2$, we have:

$\frac{dy}{dx} = e^{x^2} \cdot \frac{d}{dx}(x^2) = e^{x^2} \cdot 2x$

So: $\frac{dy}{dx} = 2x e^{x^2}$

Step 2: Find the second derivative $\frac{d^2y}{dx^2}$

Now, differentiate $\frac{dy}{dx} = 2x e^{x^2}$ with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(2x e^{x^2})$

We will use the product rule for differentiation, which states:

$\frac{d}{dx}(uv) = u'v + uv'$

Let $u = 2x$ and $v = e^{x^2}$.

1. $u' = 2$
2. $v' = \frac{d}{dx}(e^{x^2}) = 2x e^{x^2}$ (from Step 1)

Now, apply the product rule:

$\frac{d^2y}{dx^2} = 2 \cdot e^{x^2} + 2x \cdot 2x e^{x^2}$ $\frac{d^2y}{dx^2} = 2e^{x^2} + 4x^2 e^{x^2}$ $\frac{d^2y}{dx^2} = e^{x^2}(2 + 4x^2)$

Step 3: Substitute into the given equation

The equation we want to verify is:

$\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} - 2y = 0$

Substitute $\frac{d^2y}{dx^2}$, $\frac{dy}{dx}$, and $y$ into the equation:

$\frac{d^2y}{dx^2} = e^{x^2}(2 + 4x^2)$ $\frac{dy}{dx} = 2x e^{x^2}$ $y = e^{x^2}$

So, the equation becomes:

$e^{x^2}(2 + 4x^2) - 2x(2x e^{x^2}) - 2e^{x^2} = 0$

Simplify each term:

$e^{x^2}(2 + 4x^2) - 4x^2 e^{x^2} - 2e^{x^2} = 0$

Distribute $e^{x^2}$ through the first term:

$2e^{x^2} + 4x^2 e^{x^2} - 4x^2 e^{x^2} - 2e^{x^2} = 0$

Notice that the $4x^2 e^{x^2}$ terms cancel out:

$2e^{x^2} - 2e^{x^2} = 0$

Finally:

$0 = 0$

Conclusion:

The equation $\frac{d^2y}{dx^2} - 2x\frac{dy}{dx} - 2y = 0$ is satisfied for $y = e^{x^2}$.

Would you like any further explanation or details on any part of this solution?

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5 Related Questions:

1. What happens to the equation if $y = e^{kx^2}$ for some constant $k$?
2. How do you solve a differential equation of the form $y'' - 2xy' - 2y = 0$ generally?
3. Can this technique be applied to other exponential functions like $y = e^{x^3}$?
4. What is the significance of the chain rule in solving differential equations?
5. How would the solution change if a different base, such as $a^x$, is used instead of $e^x$?

Tip: Always remember to apply the product and chain rules correctly when differentiating composite functions. It helps to label parts of the function (like $u$ and $v$) to keep the work organized.