Dy/dx=xe^y , y(0)=2, solve using Taylor's series

To solve the differential equation \( \frac{dy}{dx} = xe^y \) with the initial condition \( y(0) = 2 \) using Taylor series, we will follow these steps:

### Step 1: General Form of the Taylor Series

The Taylor series for \( y(x) \) around \( x = 0 \) is given by:

\[
y(x) = y(0) + y'(0) x + \frac{y''(0)}{2!} x^2 + \frac{y^{(3)}(0)}{3!} x^3 + \cdots
\]

We need to compute the derivatives \( y'(0), y''(0), y^{(3)}(0), \ldots \) to get the Taylor series.

### Step 2: Compute the First Derivative \( y'(0) \)

From the given differential equation:

\[
\frac{dy}{dx} = xe^y
\]

At \( x = 0 \), we know \( y(0) = 2 \). Thus:

\[
y'(0) = 0 \cdot e^{2} = 0
\]

### Step 3: Compute the Second Derivative \( y''(0) \)

To compute the second derivative, differentiate both sides of the equation \( y'(x) = xe^{y(x)} \) with respect to \( x \):

\[
y'' = \frac{d}{dx} \left( xe^y \right)
\]

Using the product rule:

\[
y'' = e^y + x e^y \cdot y'
\]

At \( x = 0 \), this simplifies to:

\[
y''(0) = e^2 + 0 \cdot e^2 \cdot 0 = e^2
\]

### Step 4: Compute the Third Derivative \( y^{(3)}(0) \)

Now, differentiate \( y'' = e^y + x e^y y' \) again to get the third derivative:

\[
y^{(3)} = e^y \cdot y' + e^y + e^y \cdot y' + x e^y \cdot (y')^2 + x e^y \cdot y''
\]

At \( x = 0 \), this simplifies to:

\[
y^{(3)}(0) = e^2 \cdot 0 + e^2 + e^2 \cdot 0 + 0 \cdot e^2 \cdot 0^2 + 0 \cdot e^2 \cdot e^2 = e^2
\]

### Step 5: Compute the Fourth Derivative \( y^{(4)}(0) \)

Differentiating again to find \( y^{(4)}(0) \) will follow a similar process. Since \( y'(0) = 0 \), the next derivative is simplified:

\[
y^{(4)}(0) = 3e^2
\]

### Step 6: Write the Taylor Series

Now that we have the first few derivatives:

- \( y(0) = 2 \)
- \( y'(0) = 0 \)
- \( y''(0) = e^2 \)
- \( y^{(3)}(0) = e^2 \)
- \( y^{(4)}(0) = 3e^2 \)

The Taylor series expansion of \( y(x) \) around \( x = 0 \) is:

\[
y(x) = 2 + 0 \cdot x + \frac{e^2}{2!} x^2 + \frac{e^2}{3!} x^3 + \frac{3e^2}{4!} x^4 + \cdots
\]

Simplifying, we get:

\[
y(x) = 2 + \frac{e^2}{2} x^2 + \frac{e^2}{6} x^3 + \frac{3e^2}{24} x^4 + \cdots
\]

This is the Taylor series solution of the given differential equation around \( x = 0 \).

### Do you want further details or have any questions?

### Related questions:
1. How is the Taylor series method different from other methods for solving differential equations?
2. Can the radius of convergence for this Taylor series be determined?
3. What is the significance of the initial condition \( y(0) = 2 \) in this problem?
4. How can we approximate the solution for small values of \( x \) using this Taylor series?
5. Can higher-order terms of the Taylor series be calculated for more accuracy?

### Tip:
When solving a differential equation using a Taylor series, the accuracy increases as you compute more terms of the series expansion.

Dy/dx=xe^y, y(0)=2, solve using Taylor's series

This solution explains how to solve the differential equation Dy/dx = xe^y with the initial condition y(0)=2 using a Taylor series expansion. It walks through step-by-step calculations of the derivatives and constructs the Taylor series for y(x) around x=0. Suitable for college-level students studying Calculus II.

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