2tdy/dt+y=0
I got e^(-ln(2t)/2)
It's wrong. Please help
well, your answer is correct, if you recall that e^lnx = x
So, you got y = 1/e^(ln(2t)/2)
= 1/e^ln(√(2t))
= 1/√(2t)
That's particular. The general solution is y = c/√t
so maybe you had a condition to solve for c.