Recent content by IRNB

there is a typo in my answer. it should read x = (1/exp(-(t^2)/2))C - 2 exp(-t^2) Thank you for your help Cyosis

i have no idea why i didnt see that. its been a really long day so now i have \int 6t exp((-3/2)t2) dt u = t2 du = 2t dt \int3 exp((-3/2)u) du = 3(-2/3) exp((-3/2)u) = -2 exp((-3/2)t2) + C x exp(-t2/2) = -2 exp((-3/2)t2) + C x = (1/exp(-t2/2))C - 2 exp(-t2/2) does this look...

Thank you cyosis for all your help. My integration skills are quite weak so please bare with me here. f' = exp(-3t2/2) using u= t2 f' = exp((-3/2)u) f = \intexp((-3/2)u) (dt/du) du f = (-2/3)exp((-3/2)u) (1/2t) f = (-1/3t)exp((-3/2)t2) g = 6t g'=6 now using integration by...

Okay I've had another go and got an answer and want to check whether its correct or not. x'-tx=6te-t2 I = \int-t dt = -t2/2 xe-t2/2=\int6te-t2e-t2/2 dt xe-t2/2 = \int6te-(3/2)t2dt now using integration by parts where f'=e-(3/2)t2 and g=6t and recalling that \inte-ax2 dx =...

oh yes, i see what you mean exp(-t^2) does not equal exp(-2t). I'll give it another go. thanks guys.

Hi Physics Math The second one was just me re-arranging the first one to get it into the form of x'+p(t)x=q(t) so that i could apply the relevant formula (under relevant equations).

i just noticed a typo where i stated what value i used for f' it should read f' = e-2-t2/2

Homework Statement give the general solution of the following equation x' = tx + 6te-t2 Homework Equations for x'+p(t)x=q(t) xeI=\intq(t)eIdt where I=\intp(t)dt integration by parts \intf'g = [fg] - \intfg' The Attempt at a Solution x'-tx=6te-2t I=\int-t dt = -t2/2...

thanks a lot LowlyPion that did help. this case is now closed. :P

Thanks guys. Perhaps my lecturer was just in hurry or something and made a mistake. can anyone help out with the first part of the question? does the negative sign simply mean that the temperature is decreasing and the rate at which it changes also decreases with time? also does anyone have...

Homework Statement Newton’s law of cooling states that the rate of change of temperature of an object is proportional to the difference between the temperature, T, of the object and that of its surroundings, Ts. Derive the solution T(t) = Ts + (T0 − Ts) e−kt, where T0 is the...

Never mind. Problem solved

sorry i posted this in the wrong section. I've posted it again in the homework section. This can be deleted if need be.

Homework Statement l = Phi*L*Gamma(a+2) b) The Sloan Digital Sky Survey (SDSS) has recently measured the following Schechter parameters in the r passband: a = -1.16 +/- 0.03, M*= -20.80 +/- 0.03, Phi* = (1.50 +/-0.13). Given that the sun has absolute magnitude M = 4.62 in the SDSS r...

l = Phi*L*Gamma(a+2) b) The Sloan Digital Sky Survey (SDSS) has recently measured the following Schechter parameters in the r passband: a = -1.16 +/- 0.03, M*= -20.80 +/- 0.03, Phi* = (1.50 +/-0.13). Given that the sun has absolute magnitude M = 4.62 in the SDSS r band, calculate the...