Question

2ty/t^2+1 -2t(2-ln(t^2+1))y'=0​

Answer 1

Answer:

(4 − t2)y′ + 2ty = 3t2, y(1) = −3. (b). (t+1)( t−3) t−2 y′ + t−3 t(t−2) y = (t+1)(t+3) t. , y(1) = 0. (c) (sint)y′ + t t−1y = ln(t − 2), y(3) = 0.