Recent content by kylerawn

I redid it with the cosine function but still can determine the proper sine term at all i understand that it is the two parts

Homework Statement Use the super position method to find the solution of: y"+6y'+8y=6sin3t 2. The attempt at a solution x^2+6x+8=6sin3t found the x values x= -2,-4 yc=Asin3t+Bcos3t y'=3Acos3t-3Bsin3t y"=-9Asin3t-9Bcos3t sin3t values (8A,-18B,-9A) A=18B-6 cos3t...

I came across another problem I can't solve dy/dx=5y+1 dy/5y+1=dx exp^{5y+1} = x but i know this is incorrect can you point out where i am going wrong?

right :) thanks forgot the negitive in front. I realize it should be -1/3 i was just going with the answer my prof gave us :) Thanks for all your help

Thanks a lot \exp^{-3x}dx \-.33exp^{-3x}

dx/e^3x = 2ydy 1/e^3x = y^2 + C y= sqrt(C+e^-3x) 4= sqrt (C+ e^-3(0)) 16 = C+1 C = 15 Thats what I keep getting I know I am messing up somehwere

I tried that but i have the correct answer if y(0) = 4 the answer to the problem is sqrt(16.33 - .33e^-3x) if i rearange so dx and dy are in the numerator I get dx/e^3x = dy/2y

Homework Statement e^3x dy/dx = 1/2y Homework Equations The Attempt at a Solution e^3x dy/dx = 1/2y e^3x/dx = 1/2ydy I can't determine the dirivitives for this equation can someone help me :)