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Apr
10
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Feb
18
comment Simple rounding question
In addition to the answers already posted, this link has a good explanation about rounding in general: cstl.syr.edu/fipse/decunit/roundec/roundec.htm.
Jan
29
awarded  Popular Question
Jan
17
asked Second order ODE near regular singular point
Jan
5
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Nov
30
accepted Solve the ODE $yy'' + (y')^2 = 0$
Nov
30
comment Solve the ODE $yy'' + (y')^2 = 0$
@Did: Yes, but the book wants me to use that $v=y'$ substitution, so I am still interested in getting that solution correct.
Nov
30
revised Solve the ODE $yy'' + (y')^2 = 0$
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Nov
30
comment Solve the ODE $yy'' + (y')^2 = 0$
@Did: I put my doubt in more detail in an "Update" at the end of the original question.
Nov
30
comment Solve the ODE $yy'' + (y')^2 = 0$
This result matches the one that I found (with absolute value bars). The answer given by the book where I found this question doesn't have absolute value signs ($\dfrac{y^2}{2}=c_1t + c_2$), which I believe is incomplete, but I'm not sure (I added an update at the end of the question regarding this doubt).
Nov
30
revised Solve the ODE $yy'' + (y')^2 = 0$
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Nov
30
revised Solve the ODE $yy'' + (y')^2 = 0$
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Nov
30
comment Solve the ODE $yy'' + (y')^2 = 0$
@JJacquelin: I added an observation at the beginning of the question to make this substitution clearer (based on the book's explanation).
Nov
30
revised Solve the ODE $yy'' + (y')^2 = 0$
added 363 characters in body
Nov
30
comment Solve the ODE $yy'' + (y')^2 = 0$
@JJacquelin: To make it clearer, the intention is that v is treated as function of y, which in turn is a function of t.
Nov
30
revised Solve the ODE $yy'' + (y')^2 = 0$
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Nov
30
revised Solve the ODE $yy'' + (y')^2 = 0$
corrected spelling.
Nov
30
comment Solve the ODE $yy'' + (y')^2 = 0$
@the_candyman: Yes; this is the method suggested by the book for solving equations with the independent variable missing (equations of the form $y'' = f(y, y')$).