grouping

posted by .

i am trying to solve for A and B....this question comes from a Differential equations...but this step goes back to basics...

Asinx-Bcosx+6Acosx+6Bsinx=3-cosx

when you group all the sinx and cosx
you get

(A+6B)sinx+(6A-B)cosx=3-cosx

you get

(A+6B)=0 and (6A-B)=-1<<<<<<<<<

i am having a hard time understanding the second part where i have the <<<< next to it...i mean shouldnt it be =-3 i really don't get that...i got this answer from the back of the book but i am trying to understand it...please help if possible...

  • PLZZZ -

    plz someone help me...if anyone knowssssssss

  • grouping -

    The point of what you are doing is finding a particular solution. Ignore the -3, get the like coefficents as in the text, solve for A,B. Then you go back and try the solution you found to see if it works in the differential equation. If it does, you are ok with a particular solution. If it does not, often that checking will suggest a modification to your solution. This method is not exact, it sometimes takes iterations.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Math

    Verify the identity . (cscX-cotX)^2=1-cosX/1+cosX _______ sorry i cant help you (cscX-cotX)=1/sinX - cosX/sinX = (1-cosX)/sinX If you square this you have (1-cosX)^2/(sinX)^2 Now use (sinX)^2 = 1 - (cosX)^2 to get (1-cosX)^2 / 1 - …
  2. Pre-Calc

    Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx + cosx/cosx)/(sinx/cos x - 1/cosx + cosx/cosx)= tanx + cosx (just working on the left side) ((sinx + 1 - cosx)/cosx)/((sinx …
  3. Trigonometry.

    ( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated side …
  4. Trig........

    I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1-sinx) multiply top and …
  5. Math/Calculus

    Please take a look at my work below and provide a good critique: Solve the differential equation using the method of undetermined coefficients or variation of parameters. y'' - 3y' + 2y = sin(x) yc(x)= c1*e^2x+c2*e^x y"-3y'+2y=sin(x) …
  6. Math - Pre- Clac

    Prove that each of these equations is an identity. A) (1 + sinx + cos x)/(1 + sinx + cosx)=(1 + cosx)/sinx B) (1 + sinx + cosx)/(1 - sinx + cosx)= (1 + sin x)/cosx Please and thankyou!
  7. maths - trigonometry

    I've asked about this same question before, and someone gave me the way to finish, which I understand to some extent. I need help figuring out what they did in the second step though. How they got to the third step from the second. …
  8. Trigonometry Check

    Simplify #3: [cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] = [cosx-((1)cosx-(0)sinx)sinx]/[cosx-((-1)cosx+(0)sinx)tanx] = [cosx-cosxsinx]/[cosx+cosxtanx] = [cosx(1-sinx]/[cosx(1+tanx] …
  9. trigonometry

    can i use factoring to simplify this trig identity?
  10. Precalculus/Trig

    I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx I don't know …

More Similar Questions