Solve the following differential equation.
(dy)/(dt) + (2ty)/(1+t²) = 4ty

a) y = ((2t² + (4/3)t⁶ + C) / (1 + t²)²)
b) y = ((2/3)(1 + t²)³ + C) / (1 + t²)²
c) y = (1(1 + t²)² + C) / (1 + t²)
d) y = ((4/3)t⁶ + C) / (1 + t²)

Question

05-06-2024
Mathematics

Answered

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Solve the following differential equation.
(dy)/(dt) + (2ty)/(1+t²) = 4ty

a) y = ((2t² + (4/3)t⁶ + C) / (1 + t²)²)
b) y = ((2/3)(1 + t²)³ + C) / (1 + t²)²
c) y = (1(1 + t²)² + C) / (1 + t²)
d) y = ((4/3)t⁶ + C) / (1 + t²)

Sagot :

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Solve the following differential equation.(dy)/(dt) + (2ty)/(1+t²) = 4tya) y = ((2t² + (4/3)t⁶ + C) / (1 + t²)²)b) y = ((2/3)(1 + t²)³ + C) / (1 + t²)²c) y = (1(1 + t²)² + C) / (1 + t²)d) y = ((4/3)t⁶ + C) / (1 + t²)

Sagot :

Other Questions

Solve the following differential equation.
(dy)/(dt) + (2ty)/(1+t²) = 4ty

a) y = ((2t² + (4/3)t⁶ + C) / (1 + t²)²)
b) y = ((2/3)(1 + t²)³ + C) / (1 + t²)²
c) y = (1(1 + t²)² + C) / (1 + t²)
d) y = ((4/3)t⁶ + C) / (1 + t²)