Question

Solve y'' - 3y' + 2y = e^2x, y(0) = -1 and y'(0) = 0.

Answer 1

You have two initial conditions and a differential equation, but do you have an initial function to work with or is it just pure integration?

Answer 2

this is all that was given to me

Answer 3

what are the instructions? find y?

Answer 4

yea. here it says solve, but what you're supposed to do is find y

Answer 5

first find the homogeneous solution, do u know how to find that?

Answer 6

@danielsokolovsky

Answer 7

@lalaly care to give a hint? I feel like I should know this myself. :(

Answer 8

ok this is a second order differential equation
you need to find the homogeneous solution and the particular solution
and the general solution is the summation of both homg. and part.

first step u do is u find the homgeneous solution
you let
y''-3y'+2y=0
and solve for y..
do u want me to show u how or do u know how to do it?

Answer 9

feels like quadratic formula to me

Answer 10

either that or factor

Answer 11

yeah u change the DE into a polynomial
let \[\huge{y=e^{ \lambda x}}\]
after substitiuting
u are left with\[\lambda^2-3 \lambda+2=0\]\[(\lambda-2)(\lambda-1)=0\]
\[\lambda=2,1\]
so homogeneous solution is\[\huge{y_h=c_1e^{2x}+c_2e^x}\]

Answer 12

Next step is to find the particular solution
and there are two methods for that, u either use (method of undetermined coeffecients) or variation of parameters

Answer 13

Well, this is daniel's problem, but I have no clue what your symbols mean so meh.

Answer 14

I did find 1 and 2 though

Answer 15

lol the symbol doesnt matter, it can be t or r or whatever letter u want it to be,,, but i was just tryin to hlep:)

Answer 16

well, you did a better job than me. by the way, how did you get your symbols to be so...large?

I've never been able to get it larger than the standard size