How do I solve the following problem and what type of equation is it? (3x^2-2y) dx+ (x^2y+2x) dy=0 y(-1) =4

Answer 1

Given: (3x² - 2y) dx + (x^2y + 2x) dy = 0, y(-1) = 4

This is a first-order nonlinear ordinary differential equation.

1) Solve for the general solution

2) Use the point (-1,4) to find the specific solution

Part 1:

Because the differential form is exact, we know

(3x² - 2y) dx + (x^2y + 2x) dy = dƒ(x,y) = 0

for some ƒ(x,y). We obtain ƒ(x,y) through anti-differentiation using partial derivatives.

∫ (3x² - 2y) dx = x³ - 2xy + C₁(y)

∫ (x^2y + 2x) dy = (1/ln x) x^2y + 2xy + C₂(x)

NOTE: ∫ a^u du = (1/ln a) a^u +C

We know that f(x,y) must be in both solution spaces.

The complete candidate solution space for f is

ƒ(x,y) = x³ + (1/ln x) x^2y + C₃

Part 2:

ƒ(-1,4) = (-1)³ + (1/ln (-1)) * (-1)²⁽⁴⁾ + C₃ = 0

NOTE: 1/ln (-1) is undefined since ln(-1) is illegal.

= -1 + Undefined * (1) + C₃ = 0

= -1 + Undefined + C₃ = 0

Therefore, the general solution is correct, however there is no specific solution that passes through point (-1,4) unless we dive into complex math.