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Solve $y'(x) = a + b \cdot y(x) \cdot x$. When $x= 0$ : $y = 0$. And $a> 0$, $b > 0$ The solution to the differential equation \( y'(x) = a + bx \cdot y(x) \) with the initial condition \( y(0) = 0 \) and given \( a > 0, b > 0 \) is: $ y(x) = \dfrac{\sqrt{2} \sqrt{\pi} a e^{\dfrac{b x^2}{2}}...