Recent content by projection

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    Asteroid Impact: Risk of Debris & Intact Objects on Earth

    Is the biggest danger of an asteroid hitting Earth the debris released from the asteroid? Looking at general articles it seems that the global risk arises from the debris released into the atmosphere which would affect climate. What would happen if no debris was released? I mean, if the...
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    Ideal Diode: Exploring Negative Voltage Drop and Its Feasibility

    I am a little unsure about why the ideal diode can't have positive voltage drop... Looking back at my notes, it just simply says that "It's not a feasible solution" and that only when voltage drop is zero or negative can there be a solution.
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    2nd order inhomo differential eq.

    i am not following... i am trying to find a particular solution... am i not supposed to find a polynomial of similar degree with unknown coefficients and finding derivates and then equating it to the regular diff equation to solve for the coefficients... trying a particular solution, i am...
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    2nd order inhomo differential eq.

    Homework Statement y'' = t^2 The Attempt at a Solution The general solution to the homogenous diff equation is y(t)= C1 + tC2 i beliee. The particular solution is where i am having trouble. the guess is of the form \alpha t^2 + \beta t + \gamma ... but taking 2 deriatives leads to...
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    Solving Lcos(y)y' + sin(y) = 0

    checking my notes, the example solution stops at the statement that "the equation (exe: tsin(y)) determines y(t) implicity"... so in this case, the solution would be tsin(y) = c (implicit form) so, unless i solve for y as a function of the other variable, i should state the solution is an...
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    Solving Lcos(y)y' + sin(y) = 0

    the exact problems i have been doing usually stop at this step... and only continue if intial conditions are provided... since no intial value are given (just says test exactness and solve), i am unsure about what more i must do.
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    Solving Lcos(y)y' + sin(y) = 0

    it doesn't say, but first i thought it was a constant, but remembered than most constants in DEs come as lamba or phi etc... so looking closer at the text and removing smudge made me see it was a t... so have i done it right? the solution is tsin(y)... hate when text don't have answers for...
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    Solving Lcos(y)y' + sin(y) = 0

    opps, i think the L is actually a t... so it is not a constant i think... so would this be the solution now? \int tcos(y)dy = tsin(y) + g(y) + C \int sin(y)dt = \int dt= tsin(y) + f(y) + C and the solution is tsin(y) ?
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    Solving Lcos(y)y' + sin(y) = 0

    Homework Statement Lcos(y)y' + sin(y) = 0 The Attempt at a Solution I separated them into their respective M and N functions. took their integrals: \int Lcos(y)dy = Lsin(y) + g(y) + C \int sin(y)dx = -cos(y) + f(y) + C there the solution would be Lsine(y) -cos(y) + C...
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    What is a piecewise-defined function and how do I write one?

    thanks, i understand...but for the third part...isn't it -2x^2+5x-4? i think u made a small error in the final calculation...
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    What is a piecewise-defined function and how do I write one?

    hi, i missed a lecture on this topic and i am really lost...there is a question that asks to write a function in the piecewise-defined form. i have no idea what that means and its not in the textbook of the course...looked everywhere online and can't find anything on it... they give the...
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    How to prove these tricky trig identities?

    I think that were are supposed to prove them in the LHS and RHS formate. so i don;t think multilpying something out is allowed...
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    How to prove these tricky trig identities?

    exam coming up...need some help with these identities for practise. prove the following: A) \frac{tan^3x}{1+tan^2x}+\frac{cot^3x}{1+cot^2x}=\frac{1-2sin^2xcos^2x}{sinxcos} B) sec^6x-tan^6x=1+3tan^2xsec^x C) cos^4x=\frac{3}{8}+\frac{1}{2}cos2x+\frac{1}{8}cos4x D)...
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    Understanding y=\frac{\sqrt{x}}{2^x} and y=lnx^{{lnx}^{{lnx}^{{lnx}^{{lnx}}}}}

    yes, i stated that in the title, sorry for not talking about it in the thread.
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