Consider the differential equation dy/dt = (e^t)(y)(lny)
What is the general solution of this equation?
I know that you have to split the integral into
dy/(y*lny) = (e^t)dt
I'm not sure how to integrate 1/(y*lny)
Please help!
What is the general solution of this equation?
I know that you have to split the integral into
dy/(y*lny) = (e^t)dt
I'm not sure how to integrate 1/(y*lny)
Please help!
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To integrate dy/(y lny) use a substitution:
u = lny
du = 1/y dy
∫ dy/(y lny) = ∫ 1/lny * 1/y dy
. . . . . . . . . = ∫ 1/u du
. . . . . . . . . = ln(u)
. . . . . . . . . = ln(lny)
u = lny
du = 1/y dy
∫ dy/(y lny) = ∫ 1/lny * 1/y dy
. . . . . . . . . = ∫ 1/u du
. . . . . . . . . = ln(u)
. . . . . . . . . = ln(lny)
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see d(ln ln y) = 1/ln y * 1/y * dy
that gives d(ln ln y) = d(e^t).
Integrate both side.
1. To see how I got the answer, just differentiate ln ln y, you will see that d (ln ln y) = dy/(y*lny) by successive differentiation rule.
Integrating both sides means
ln ln y = e^t + c
that gives d(ln ln y) = d(e^t).
Integrate both side.
1. To see how I got the answer, just differentiate ln ln y, you will see that d (ln ln y) = dy/(y*lny) by successive differentiation rule.
Integrating both sides means
ln ln y = e^t + c