- #1
asdf1
- 734
- 0
for the following question:
y``+2y`-35y=12e^(5x)+37sin5x
my problem:
yh=c1e^(-7x)+c2e^(5x)
suppose yp=c3xe^(5x)+c4sinx+c5cos5x
then yp`=c3e^(5x)+5c3xe^(5x)+5c4cos5x-5c5sin5x
so yp``=8c3e^(5x)+(-60c4-10c5)sin5x+(-60c5+10c4)
which means that 8c3=12 => c3=2/3
also from (-60c4-10c5)=37 and (10c4-60c5)=0 implies that
c4=-3/5 and c5=-1/10
so y=c1e^(-7x)+c2e^(5x)+(3/2)xe^(5x) +(3/2)sin5x-(1/10)cos5x
but the correct answer should be
y=c1e^(-7x)+c2e^(5x)+xe^(5x) +(3/2)sin5x-(1/10)cos5x
i've checked this many times, but i don't know where my calculations went wrong... :P
y``+2y`-35y=12e^(5x)+37sin5x
my problem:
yh=c1e^(-7x)+c2e^(5x)
suppose yp=c3xe^(5x)+c4sinx+c5cos5x
then yp`=c3e^(5x)+5c3xe^(5x)+5c4cos5x-5c5sin5x
so yp``=8c3e^(5x)+(-60c4-10c5)sin5x+(-60c5+10c4)
which means that 8c3=12 => c3=2/3
also from (-60c4-10c5)=37 and (10c4-60c5)=0 implies that
c4=-3/5 and c5=-1/10
so y=c1e^(-7x)+c2e^(5x)+(3/2)xe^(5x) +(3/2)sin5x-(1/10)cos5x
but the correct answer should be
y=c1e^(-7x)+c2e^(5x)+xe^(5x) +(3/2)sin5x-(1/10)cos5x
i've checked this many times, but i don't know where my calculations went wrong... :P