Asked by R
2tdy/dt+y=0
I got e^(-ln(2t)/2)
It's wrong. Please help
I got e^(-ln(2t)/2)
It's wrong. Please help
Answers
Answered by
oobleck
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.
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.
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