Differential Equation prob

Could someone solve and explain this problem

y'-4xy=x

It's first order linear. Use the integrating factor e^[∫ -4x dx] = e^(-2x²). Multiply through by the integrating factor and note that the left side collapses as a product rule.

e^(-2x²)(y ' - 4xy) = x e^(-2x²) ==>

d/dx [e^(-2x²)y ] = x e^(-2x²).

Integrate

e^(-2x²)y = -e^(-2x²)/4 + C.

Isolate y by multiplying through by e^(2x²), the reciprocal of e^(-2x²).

y = -1/4 + Ce^(2x²).