Question

Find the equation of the curve which passes through the point (1 , 4) and is such that : dy÷dx =2x^2 +3x +2

Answer 1

Let us rewrite the differential equation in form $dy = (2 x^2 + 3 x + 2 ) dx$, and integrate both sides: $y = \int dy = \int (2 x^2 + 3 x + 2 ) dx =\frac{2}{3} x^3 + \frac{3}{2} x^2 + 2 x + C$.

Substituting the boundary condition $y(1) = 4$ into last equation, obtain $4 = \frac{25}{6} + C$, from where $C = -\frac{1}{6}$. Hence, the solution of given differential equation is $y(x) = \frac{2}{3}x^3 + \frac{3}{2} x^2 + 2 x - \frac{1}{6}$ .