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 y''+5y=x+2.

Solve first for y''+5y=0.

Let y=Asin(x√5)+Bcos(x√5), where A and B are constants, then:

y'=(A√5)cos(x√5)-(B√5)sin(x√5),

y''=-5Asin(x√5)-5Bcos(x√5)=-5y.

Therefore y''+5y=0.

Let y=ax+b, then y''=0 and y''+5y=5ax+5b.

Therefore 5ax+5b≡x+2.

Matching coefficients, a=⅕, b=⅖.

Now combine the two solutions for y:

y=Asin(x√5)+Bcos(x√5)+(x+2)/5 is the whole solution.

ago by Top Rated User (813k points)

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