Recent content by enosthapa

Oh yes thankx... but can I do ln(e^-x) = -x lne =-x

I would make sense but this is what i did ln(e^-x) = 1/(lne^x) =1/(x.lne) =1/x

Homework Statement Differentiate y=e-xsin(3x) 2.The attempt at a solution With Product rule y'=-e-xSin(3x)+3e-xCos(3x) And Logarithm way Taking log both sides lny= lne-x+ln(Sin3x) lny= 1/x+ln(Sin3x) 1/y*dy/dx=-x-2+3(Cos3x/Sin3x) y'=-(e-xSin3x)/x2+3e-xCos3x

I don't have the answer which could have been helpful. But thanks anyway.

Thanks a lot but I am still confused As far as i know, if there is exponential on RHS (ekx) and if 'k' is simple root of the auxiliary equation which it is in this case, then we try a.x.ek.x i.e a.x.e-x but the question has polynomial in addition so did you mean i have to add the 'PI' I...

y''+5y'+4y=x^2+2e^(-x) Auxiliary roots: m1=-1 m2=-4 Using trial please cant seem to work out PI with polynomial and exponential together.

I have done some calculations to find out the forces but i am confused with the middle section. Can someone tell me if figure 2 is correct or not?Thank you

Could you help me find out what force (tension/compression) do the truss members have please? I have done most of them but the middle section with "X" trusses are making it dificult to calculate the forces the carry...

In general, in a truss diagram the forces like load, reaction force are act on joints but in this case the load is in the middle of two joints. How can I find forces on members in this case? Your help is really appreciated. Thank you for reading