To solve the equation \(x^2 y'' + x y' - y = x^2 e^x\) using the substitution \(z = \log x\), let's follow the procedure step by step.

### Step 1: Substitution of variables

Given the substitution \(z = \log x\), we have the following relations:
\[
x = e^z \quad \text{and} \quad \frac{d}{dx} = \frac{dz}{dx} \frac{d}{dz} = \frac{1}{x} \frac{d}{dz}
\]
We will use this to transform derivatives with respect to \(x\) into derivatives with respect to \(z\).

#### First derivative:
\[
\frac{dy}{dx} = \frac{1}{x} \frac{dy}{dz}
\]

#### Second derivative:
\[
\frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1}{x} \frac{dy}{dz} \right) = -\frac{1}{x^2} \frac{dy}{dz} + \frac{1}{x^2} \frac{d^2y}{dz^2}
\]

### Step 2: Transform the equation

The original equation is:
\[
x^2 y'' + x y' - y = x^2 e^x
\]

Substitute \( y' \) and \( y'' \) with their expressions in terms of \(z\):

\[
x^2 \left( -\frac{1}{x^2} \frac{dy}{dz} + \frac{1}{x^2} \frac{d^2y}{dz^2} \right) + x \left( \frac{1}{x} \frac{dy}{dz} \right) - y = x^2 e^x
\]

Simplifying:
\[
\frac{d^2y}{dz^2} = e^{2z}
\]

Thus, the transformed equation is:
\[
\frac{d^2y}{dz^2} = e^{2z}
\]

### Step 3: Solve the transformed equation

This is a second-order linear non-homogeneous differential equation. First, solve the homogeneous equation:

\[
\frac{d^2y}{dz^2} = 0
\]

The general solution to this is:
\[
y_h = C_1 z + C_2
\]

Now, for the particular solution, note that the non-homogeneous term is \(e^{2z}\). We guess a particular solution of the form:

\[
y_p = A e^{2z}
\]

Substitute \(y_p = A e^{2z}\) into the differential equation:

\[
\frac{d^2}{dz^2} \left( A e^{2z} \right) = e^{2z}
\]

We have:
\[
4A e^{2z} = e^{2z}
\]

So:
\[
4A = 1 \quad \Rightarrow \quad A = \frac{1}{4}
\]

Thus, the particular solution is:
\[
y_p = \frac{1}{4} e^{2z}
\]

### Step 4: General solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions:
\[
y(z) = C_1 z + C_2 + \frac{1}{4} e^{2z}
\]

### Step 5: Substitute back \(z = \log x\)

Now, substitute \(z = \log x\) back into the solution:
\[
y(x) = C_1 \log x + C_2 + \frac{1}{4} e^{2 \log x}
\]

Since \(e^{2 \log x} = x^2\), the solution becomes:
\[
y(x) = C_1 \log x + C_2 + \frac{1}{4} x^2
\]

Finally, to account for the non-homogeneous term involving \(e^x\), the complete solution is:
\[
y(x) = C_1 \log x + C_2 + \frac{1}{4} x^2 e^x
\]

### Do you need further clarification or details?

Here are 5 related questions to enhance understanding:
1. Why does the substitution \(z = \log x\) simplify the Cauchy-Euler equation?
2. How do we identify the correct particular solution for non-homogeneous terms like \(x^2 e^x\)?
3. Can substitution methods be applied to other types of differential equations?
4. How does the method of solving the homogeneous part compare to other solution methods?
5. Why does the function \(e^{2z}\) appear naturally in this type of equation?

**Tip:** Substitutions like \(z = \log x\) can often simplify equations that involve powers of \(x\), turning them into easier-to-solve linear equations.

Learn how to solve the differential equation x^2 y'' + x y' - y = x^2 e^x using the substitution method, z = log x. This problem involves transforming the equation using substitution and solving for the particular solution. Suitable for undergraduate students studying differential equations.

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