Question

if you have knowledge of differential equations please answer my question...thank you

Solve the particular solution of this differential equation: y''+9y=15.25cos(3.1x)

Answer 1

Write p(D) = D^2 + 9. Then the particular solution of the equation is
y_p = 15.25/p(3.1) cos(3.1x). Substitute and you'll see this works.

Answer 2

may i ask how u arrived at p(D)=D^2+9

Answer 3

oh wait a minute thats the complementary side

Answer 4

sir please explain what you did...thank you

Answer 5

Yes, you see that characteristic equation of the LHS. I.e., if you used a trial solution of exp(ax), you would end up with p(a) as a factor on the LHS.

It is also the case that if D is the differential operator, then LHS = p(D)y.

Answer 6

ok the complementary solution is yc=c1cos3x+c2sin3x

Answer 7

can you show me what i should use as my yp?

Answer 8

the way i arrived at the yx is you notice homogenous equation is m^2+9=0 so the m1=3i and m2=-3i, therefore yc=c1e^(3i)+c2e^(-3i) and if you use euler's identity you get yc=c1cos3x+c2sin3x

Answer 9

now the way i am doing this i am letting yp=Acos3.1x+Bsin3.1x but the solution repeats so i multiply it by another x and i get yp=Axcos3.1x+Bxsinx3.1x but sir this does not work somehow!

Answer 10

Just what I said. Watch the first 20 minutes of this lecture:
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-13-finding-particular-sto-inhomogeneous-odes/

Answer 11

ok sir

Answer 12

Sorry, I should have said the particular solution is
y_p(x) = 15.25/p(3.1i) cos(3.1x)

I need that i in there. Calculating out a little, p(3.1i) = -3.1^2 + 9 = -0.61

So unsimplified y_p(x) = -15.25/0.61 cos(3.1x)

Substitute this equation back into your differential equation and you will indeed see it is a solution of the inhomogeneous equation.

Answer 13

thanks for your help appreciate it but unfortunately we are suppose to use method of undetermined coefficients to solve this problem