First, we determine the general solution of the associated homogeneous ODE y'(x) = p(x) \cdot y(x).

It is given by y(x) = K \cdot e^{P(x)}.

Here, K is a constant and \displaystyle P(x) \in \int p(x) dx is an antiderivative of p.

For the case at hand, we have \displaystyle P(x) \in \int p(x) dx = \int \left( 2x - A+B \right) dx = x^2 -A+Bx+K.

The integration constant K is determined by the initial condition y(0) = 0.

With the method of variation of constants, the general solution of the inhomogeneous ODE is \displaystyle y(x) = \left( \int e^{-P(x)} \cdot q(x) dx + C \right) e^{P(x)}.

We calculate e^{-P(x)} \cdot q(x) = e^{-x^2 +A+Bx} e^{x^2 - A+Bx +A*B} = e^{A*B}

and \displaystyle \int e^{-P(x)} \cdot q(x) dx = \int e^{A*B} dx = x e^{A*B} + C.

In the general solution \displaystyle y(x) = \left( x e^{A*B} + C \right) e^{P(x)}, we find the constant C by

y(0) = 0 = \left( 0 \cdot e^{A*B} + C \right) e^{P(0)} = C.

Thus, the solution of the IVP is \displaystyle y(x) = x e^{A*B} e^{P(x)} = e^{A*B} x e^{x^2 -A+Bx}.

Substitution gives {\color{orange} y(C)} = e^{A*B} \cdot C \cdot e^{C^2 -A+B \cdot C} = e^{A*B} \cdot C \cdot e^{-A*B} = C .