Question

find general solution:

y''+9y=2cos3x

Answer 1

Hmm, it's been a while since i did these type of equations, give me a minute.

Answer 2

can you find the complimentary solution?

Answer 3

@weedle, do you know how to find the complimentary solution or can you do that?

Answer 4

Zimmah, is this basic differential equations or upper division?

Answer 5

basic, it's just a second order linear

Answer 6

I see, we haven't reached this point in my class yet, which is why I was wondering lol

Answer 7

It's not as hard as it looks really, you just need to find a function, kinda like a piece of a puzzle, that becomes, in this case 2cos3x after you differentiate it a number of times

Answer 8

Because in this case, the function must ultimately become a cosine function, the original function is either a sine, or a cosine, Because if you differentiate a sine or cosine function you will be left with either. Any other fuction won't result in a cosine function

Answer 9

So what we are looking for, is a function y(x) that if differentiated twice, and multiplied 9 times, results in 2cos(3x)

Answer 10

is that how most second order differentials are solved?

Answer 11

it's one way of solving them, but i wouldn't say it's the most common way, there's a ton of ways to solve these problems.

Answer 12

were still on first order and now we are moving on to linear algebra so I guess ill get to second order later on in the semester

Answer 13

in this particular case you can solve it by assuming \[\Large e^{\lambda x}\] as a solution to y.

this would give \[\Large \frac{ \delta^2 y }{ \delta^2 x }(e^{\lambda x})+9e^{\lambda x}=0\]

solving that gives \[\lambda = 3i~or~-3i\]

Answer 14

then since y(x)=\[\Large y(x)=y_{1}+y_{2}=c_{1}e^{(3i)x}+\frac{ c_{2} }{ e^{(3i)x} }\]

by using \[\Large e^{a+bi \beta}= e^{\alpha}\cos(\beta)+ie^{\alpha}\sin(\beta)\]

you get \( y(x)=c_{1}(cos(3x)+isin(3x))+c_{2}(cos(3x)-isin(3x)) \)

Answer 15

regroup terms to get \[y(x)=(c_{1}+c_{2})\cos(3x)+i(c_{1}-c_{2})\sin(3x)\]

now define \( (c_{1}+c_{2})~as~c_{3}~and~i(c_{1}-c_{2})~as~c_{4}\)

Answer 16

\( y(x)=c_{3}cos(3x)+c_{4}sin(3x) \)

now use this y in the original equation to find \(c_{3} \) and \(c_{4}\) and that should be it

Answer 17

oh thats not too bad. Nice explanation btw

Answer 18

That's the complementary solution,
ie the solution to y''+9y=0

to get the general solution you'll need to add this the the particular solution,