We are asked to solve the second-order linear non-homogeneous differential equation $y'' - y' = 2 - 2x$ with initial conditions $y(0) = 1$ and $y'(0) = 1$.

1. Problem Description

We are asked to solve the second-order linear non-homogeneous differential equation $y'' - y' = 2 - 2x$ with initial conditions $y(0) = 1$ and $y'(0) = 1$.

2. Solution Steps

First, we solve the homogeneous equation $y'' - y' = 0$.

The characteristic equation is $r^2 - r = 0$, which factors as $r(r-1) = 0$.

Thus, the roots are $r_1 = 0$ and $r_2 = 1$.

The general solution to the homogeneous equation is $y_h(x) = c_1 e^{0x} + c_2 e^{1x} = c_1 + c_2 e^x$.

Now we find a particular solution to the non-homogeneous equation $y'' - y' = 2 - 2x$.

Since the right-hand side is a polynomial of degree 1, we try a solution of the form $y_p(x) = Ax^2 + Bx + C$. However, since $r=0$ is a root of the characteristic equation, we must multiply by $x$. So, let $y_p(x) = Ax^2 + Bx$. Then, $y_p'(x) = 2Ax + B$ and $y_p''(x) = 2A$. Substituting these into the given differential equation:

$2A - (2Ax + B) = 2 - 2x$.

$2A - 2Ax - B = 2 - 2x$.

Equating coefficients, we have:

$-2A = -2$, so $A = 1$.

$2A - B = 2$, so $2(1) - B = 2$, which means $B = 0$.

Thus, $y_p(x) = x^2$.

The general solution is $y(x) = y_h(x) + y_p(x) = c_1 + c_2 e^x + x^2$.

Now we apply the initial conditions.

$y(0) = c_1 + c_2 e^0 + 0^2 = c_1 + c_2 = 1$.

$y'(x) = c_2 e^x + 2x$.

$y'(0) = c_2 e^0 + 2(0) = c_2 = 1$.

Since $c_2 = 1$, we have $c_1 + 1 = 1$, so $c_1 = 0$.

Therefore, $y(x) = 0 + 1 e^x + x^2 = e^x + x^2$.

3. Final Answer

$y(x) = e^x + x^2$