for the following question:(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: Linear differential equaton

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