If y= e^x^2 show that d^2y / d^2x - 2x(dy/dx) - 2y = 0

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?

---

**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.

Learn how to verify the differential equation d^2y/dx^2 - 2x(dy/dx) - 2y = 0 for y = e^x^2 using step-by-step differentiation and substitution. Explore the application of exponential functions and the chain rule in differential equations.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation