engineering recuitment AAI JE Technical Mock Test 2020-21 Engineering Mathematics Differential Equations
The solution of the differential equation
t2y’ – y2 – yt = 0, t > 0 is
1
\(y = \frac{{ - 1}}{{\ln t + C}}\)
2
\(y = \frac{{ - t}}{{\ln t + C}}\)
3
\(y = \frac{1}{{\ln t + C}}\)
4
\(y = \frac{t}{{\ln t + C}}\)