(3y^2 + 2)dy/dx = 1 and y(-1) = 1. what is x when y = 2? x =

Answer

Explanation:

Step1: Separate variables

Separate the variables in the differential equation ((3y^{2}+2)\frac{dy}{dx}=1) to get ((3y^{2}+2)dy = dx).

Step2: Integrate both sides

Integrate (\int(3y^{2}+2)dy=\int dx). Using the power - rule of integration (\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C) ((n\neq - 1)), we have (y^{3}+2y=x + C).

Step3: Find the constant C

Use the initial condition (y(-1)=1). Substitute (x=-1) and (y = 1) into (y^{3}+2y=x + C). Then (1^{3}+2\times1=-1 + C), which simplifies to (1 + 2=-1 + C), and (C=4). So the equation is (y^{3}+2y=x + 4).

Step4: Solve for x when y = 2

Substitute (y = 2) into (y^{3}+2y=x + 4). We get (2^{3}+2\times2=x + 4), which is (8 + 4=x + 4). Then (x=8).