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...
plz someone help me...if anyone knowssssssss
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.
To solve for A and B, we have the equation:
(A+6B)sinx + (6A-B)cosx = 3 - cosx
Grouping the sinx and cosx terms yields:
(A+6B)sinx + (6A-B)cosx = 3 - cosx
To solve this equation, we equate the coefficients of the sinx and cosx terms to the coefficients on the right-hand side of the equation.
Comparing the coefficients of sinx, we have:
A + 6B = 0 ---- (1)
Comparing the coefficients of cosx, we have:
6A - B = -1 ---- (2)
Now, let's focus on the second part where you have the "<<<<<" symbol. You mentioned that you expected the equation to be equal to -3 instead of -1.
Looking at equation (2), we have:
6A - B = -1
To isolate B, we can add B to both sides of the equation:
6A = B - 1
Next, we add 1 to both sides:
6A + 1 = B
So the correct equation for B should be B = 6A + 1, not B = 6A - 1.
Therefore, the correct set of equations to solve for A and B are:
A + 6B = 0
6A + B = 1
Solving these equations simultaneously will give you the values for A and B.