Solve $y''=12x^2$ where $y'(0) = 15$ and $y(0) = 13$ Itegrate:$y' = int12x^2dx = 4x^3 + c_1$Using the first condition $y'(0) = 15$:$4(0)^3 + c_1 = 15$$c_1 = 15$Therefore, the differential equation is:$y' = 4x^3 + 15$Integrate again:$y = int(4x^3 + 15)dx = x^4 + 15x + c_2$Using the second condition $y(0) = 13$:$0^4 + 15*0 + c_2 = 13$$c_2 = 13$The answer is:$y = x^4 + 15x + 13$