Question
Solve the differential equation: dy/dx = (3x^2) / (y+1) with initial conditions x = -1, y = 2
I have:
∫y + 1 dy = ∫3x^2 dx
(y^2 / 2) + y = x^3 + C
(2^2 / 2) + (2) = (-1^3) + C?
4 = -1 + C
5 = c
(y^2 / 2) + y = x^3 + 5
Is this correct?
I have:
∫y + 1 dy = ∫3x^2 dx
(y^2 / 2) + y = x^3 + C
(2^2 / 2) + (2) = (-1^3) + C?
4 = -1 + C
5 = c
(y^2 / 2) + y = x^3 + 5
Is this correct?
Answers
Steve
You should probably solve for y using the quadratic formula. It comes out
y = √(2x^3+11) - 1
y = √(2x^3+11) - 1
Omar
But other than solving for y, is it correct?