the formula is A cosx + b sinx
yp: A cos x +b sinx
y'p: -A sinx +bcosx
y"p: -A cosx -b sinx
(A-A+b)cosx +(-A+b-b)sinx= sinx
B cosx + -Asinx = sinx
A=-1
b=0
Yp: -1 cosx
y'p : sinx
y"p : cosx
now insert them back in gives you y"+y'+y= sinx...
How about this problem did i do this correct ?
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L(y)= y''-3y'+2y find L (e^x) and L (Xe^x)
Hence solve the equation y''-3y'+2y=3e^x
This is how I started to work out this problem.
y1=e^x...
I also tried Yp: A cosx + b cos x , after taking the derivatives they seem to cancel out and leave :A sinx + b cosx = sinx A=1, and B=0 it does not seem to work ? I must be missing something or working it wrong?
I need help solving this equation y''+y'+y= sinx
I know it looks simple but It seems to be getting sticky! I have been trying to solve it using variation of perimeters,maybe there is a quicker way? If anyone can help please...:bugeye: