y"-4y=12x y(0)=4 and y'(0)=1?
in Calculus Answers by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

y"-4y=12x.

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.

y=3e^2x+e^-2x-3x.

by Top Rated User (823k points)

Related questions

Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
86,111 questions
92,095 answers
2,244 comments
23,895 users