use the power series expansion method to find the solution, up to and including the term v in x^(4), of the following differential equation with the initial values: (d^(2)y)/(dx^(2))-4(dy)/(dx)+e^(x)y=x^(2),y(0)=1,y^(')(0)=1

1. **Assume a power series solution:**

y(x) = \sum_{n=0}^{\infty} a_n x^n

2. **Find the derivatives of \( y(x) \):**

y'(x) = \sum_{n=1}^{\infty} a_n n x^{n-1}
y''(x) = \sum_{n=2}^{\infty} a_n n (n-1) x^{n-2}

3. **Substitute the series and its derivatives into the differential equation:**

\sum_{n=2}^{\infty} a_n n (n-1) x^{n-2} - 4 \sum_{n=1}^{\infty} a_n n x^{n-1} + \sum_{n=0}^{\infty} a_n e^x x^n = x^2

4. **Equate the series on both sides of the equation:**

- Matching coefficients for \( x^0 \):
a_0 = 1

- Matching coefficients for \( x^1 \):
a_1 = 1

- Matching coefficients for \( x^2 \):
2a_2 - 4a_1 = 0 \implies a_2 = 2

- Matching coefficients for \( x^3 \):
6a_3 - 4a_2 = 0 \implies a_3 = \frac{4}{3}

- Matching coefficients for \( x^4 \):
12a_4 - 8a_3 + a_0 = 1 \implies 12a_4 - 8 \left( \frac{4}{3} \right) + 1 = 1 \implies a_4 = \frac{5}{12}

5. **Combine the coefficients to write the power series solution:**

y(x) = 1 + x + 2 x^2 + \frac{4}{3} x^3 + \frac{5}{12} x^4

Therefore, the solution up to and including the term involving \( x^4 \) is:

y(x) = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{5x^4}{24}