Answer to Question #190528 in Differential Equations for Samiullah Jan

Question

2ty/t^2+1 -2t(2-ln(t^2+1))y'=0​

Answer 1

Given,

$(4+t^2)\dfrac{dy}{dt}+2ty=4t$

$\\[9pt] \dfrac{dy}{2-y}=\dfrac{2tdt}{4+t^2}$

Integrating Both the sides-

$-ln(2-y)=ln(t^2+4)+ln c_1 \\[9pt] or c=\dfrac{2-y}{t^2}{4}\\[9pt] So ,y=2-\dfrac{c}{t^2+4}=\dfrac{2t^2+8-c}{t^2+4}$