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The differential equation given is:

\[
\frac{dy}{dx} = x^2 e^{y + 2x}
\]

with the initial condition \( y = 0 \) when \( x = 0 \). The goal is to solve this differential equation and express \( y \) in terms of \( x \).

### Step 1: Rearranging the Equation
The equation can be rearranged to isolate \( y \) and \( x \) terms:

\[
\frac{dy}{e^y} = x^2 e^{2x} \, dx
\]

### Step 2: Integrate Both Sides
To solve this, integrate both sides separately:

\[
\int \frac{1}{e^y} \, dy = \int x^2 e^{2x} \, dx
\]

For the left-hand side:

\[
\int \frac{1}{e^y} \, dy = \int e^{-y} \, dy = -e^{-y}
\]

For the right-hand side, integration by parts or using substitution methods might be required. Let's proceed with that step.

### Right-hand Side Integration:
To solve \( \int x^2 e^{2x} \, dx \), we can use the method of integration by parts. Let:

- \( u = x^2 \) and \( dv = e^{2x} \, dx \)
- Thus, \( du = 2x \, dx \) and \( v = \frac{1}{2} e^{2x} \)

Now applying integration by parts:

\[
\int x^2 e^{2x} \, dx = x^2 \cdot \frac{1}{2} e^{2x} - \int 2x \cdot \frac{1}{2} e^{2x} \, dx
\]

\[
= \frac{x^2}{2} e^{2x} - \int x e^{2x} \, dx
\]

We again apply integration by parts on \( \int x e^{2x} \, dx \):

- \( u = x \) and \( dv = e^{2x} \, dx \)
- Thus, \( du = dx \) and \( v = \frac{1}{2} e^{2x} \)

\[
\int x e^{2x} \, dx = x \cdot \frac{1}{2} e^{2x} - \int \frac{1}{2} e^{2x} \, dx = \frac{x}{2} e^{2x} - \frac{1}{4} e^{2x}
\]

Thus:

\[
\int x^2 e^{2x} \, dx = \frac{x^2}{2} e^{2x} - \left(\frac{x}{2} e^{2x} - \frac{1}{4} e^{2x}\right) = \left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) e^{2x}
\]

### Step 3: Combining and Applying the Initial Condition
Now we have:

\[
-e^{-y} = \left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) e^{2x} + C
\]

Applying the initial condition \( x = 0 \) and \( y = 0 \):

\[
-e^{0} = \left(\frac{0^2}{2} - \frac{0}{2} + \frac{1}{4}\right) e^{0} + C \implies -1 = \frac{1}{4} + C \implies C = -\frac{5}{4}
\]

Thus, the solution is:

\[
-e^{-y} = \left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) e^{2x} - \frac{5}{4}
\]

### Step 4: Solving for \( y \)
To find \( y \), isolate \( e^{-y} \):

\[
e^{-y} = -\left(\left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) e^{2x} - \frac{5}{4}\right)
\]

Taking the natural logarithm on both sides to find \( y \):

\[
y = -\ln\left(\left(\frac{x^2}{2} - \frac{x}{2} + \frac{1}{4}\right) e^{2x} - \frac{5}{4}\right)
\]

This is the final expression for \( y \) in terms of \( x \).

Would you like more details on the steps, or do you have any questions?

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Here are five related questions:

1. How does the integration by parts method work, and when should it be used?
2. Why is an integrating factor not applicable in solving this differential equation?
3. How do we interpret the natural logarithm of an expression in the context of differential equations?
4. What would happen if a different initial condition were given instead of \( y = 0 \) at \( x = 0 \)?
5. Could we solve this differential equation numerically if we couldn't find a closed form?

**Tip:** In solving differential equations, always check if you can separate variables before resorting to more complex techniques like integration by parts or substitution.

The variables x and y satisfy the differential equation dy/dx = x^2 * e^(y + 2x) and it is given that y = 0 when x = 0. Solve the differential equation and obtain an expression for y in terms of x.

This article provides a detailed solution to the differential equation dy/dx = x^2 * e^(y + 2x) with the initial condition y = 0 when x = 0. Learn about separation of variables, integration by parts, and how to apply initial value conditions.

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