please solve this linear differential equation and show your steps: y''+9y=15.25cos3.1x
this is a second order differential equation - is the y" correct - might it not be y' ?
yes it is a second order diff. equation
ok
the complementary solution is m^2+9=0, so you get m1=3i and m2=-3i so the yc=c1cos3x+c2sin3x
i am having trouble finding the particular solution...
ok the PI is of the form C cos 3.1x + D sin 3.1x
just curious have u taken a course in differentials?
ahah - i did them a long time ago! i'm retired now and a little rusty i must admit.
haha i just want to inform that with your method the complementary solution equals the particular solution which can not happen so in this case the particular solution has to be multiplied by another x to make it look different then we find the values of C and D and that is where i am stuck....
this problem is fairly advanced...i am really good at diff eq's but i been tryin to tackle this for past couple hours...and its really annoyin me
right - well i have some time on my hands later on -so i'll come back to you. thats the best i can do i'm afraid - maybe someone else will be able to help meantime.
no problem
hello
hi - i'm beaten by this problem mark o - any ideas ?
wolfram gives the solution as y(x) = c1cos3x + c2sin3x - 1.69298x + 0. getting the PI is the hard part
try Yp=Acos 3.1x + Bsin 3.1x Y'p=-3.1Asin 3.1x+ 3.1Bcos3.1x Y''p=-9.61A cos3.1x-9.61B sin3.1x now sub in the original eq y''+9y=15.25cos3.1x -9.61A cos3.1x-9.61B sin3.1x +(9Acos 3.1x + 9Bsin 3.1x)=15.25cos3.1x -0.61A cos3.1x-0.61B sin3.1x =15.25cos3.1x -0.61A=15.25 , A=15.25/-0.61=-25 and B=0 therefore Yp=-25cos3.1x Y=Yc+Yp Yc=C1cos 3x+C2 sin3x--25cos3.1x ........G.S.
i thought about doing it this way but m eeng said it was'nt valid (see above) but the CF is not = PI as the cos value is different. but where did 1.69298 x come from ?
hello ....ah is that the wolfram results?
oops sorry double negative..lol Y=C1cos 3x+C2 sin3x-25cos3.1x ........G.S.
i need to review my solution .....
sorry gotta go - be back soon
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