y"-4y=12x y(0)=4 and y'(0)=1?
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Let y=p+q where p and q are functions of x, then y"=p"+q".

So p"+q"-4p-4q=12x.

If p"-4p=0 can be solved we only have to solve q"-4q=12x.

p"-4p=0 has the characteristic equation (m-2)(m+2)=0 and we use the roots to suggest what p might be:

p=Ae^2x+Be^-2x; p'=2Ae^2x-2Be^2x; p"=4Ae^2x+4Be^2x=4p.

Now for q. Since q"=0, q'=a, a constant, and q=ax+b.

q"-4q must be equal to 12x so -4ax-4b=12x, making b=0 and a=-3.

Therefore q=-3x and y=p+q=Ae^2x+Be^-2x-3x; y'=2Ae^2x-2Be^2x-3.

y(0)=4=A+B; y'(0)=1=2A-2B-3, A-B=2. So 2A=6, A=3 and B=1.


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