Calculus 3 (Differential Equation)

Question

Prove that each equation is a solution of the given differential equation:

Y^-3 =x³(3e^x+C) ; xy'+y+x⁴y⁴e^x=0

Answer 1

By differentiation we have: - 3y'/y⁴ = 3x²(3e^x + C) + x³·3e^x; - y'/y⁴ = x²[1/(x³y³)] + x³e^x; xy' + y + x⁴y⁴e^x = 0.