Question

How do I solve this y′′−3y′+2y=3e^(-x)-10cos(3x) y(0)=5, y'(0)=-2?
i can solve the first part of the equation by using the homogeneous solutions of c1e^x+c2e^2x+yp=0 and but when i try solving the yp i get no were.

Answer 1

for non-homogenous part, you have 2terms, right?, so solve first term first
It it 3e^(-x) . That is y'' -3y'+2y = 3e^(-x)
the same with the second term, then combine them all.

Answer 2

for homogenous part, why don't you replace y(0) and y'(0) to find C1, C2?

Answer 3

i thought i had to solve for a general soultion before i input my iv

Answer 4

oh if i seperate them would i be doing the undetermined coefficients methods for both of them?

Answer 5

I think you misunderstand the method!!
You have 2 parts: homogenous y"-3y'+2y =0 , that will give you 2 homogeneous solutions, let call y_g1 and y_g2
non homogenous part , that I showed you above, that will give you some other solutions. then. combine them all

Answer 6

i think im lost

Answer 7

For homogenous part: Characteristic equation is r^2 -3r +2 = 0 , r = 2 and r =1
that gives you the solutions: y_g1 = C1 e ^2x and y_g2 = C2 e^x

Answer 8

Now, plug y(0) = 5 and y'(0) = -2
you have C1, C2, what are they?

Answer 9

i got my homogeneous solution but for the second part the 3e^(-x)-10cos3x i cant solve

Answer 10

c1=-7 and c2=12

Answer 11

you use particular integrals for them

Answer 12

ok, let amorfide cooks this meal. :)

Answer 13

ok. When you mean particular integrals do you mean when you set cx=ae^(-x)cos3x+be^(-x)sin3x and take two derivatives of it to get c'' and c'?

Answer 14

and then pop those results into the original equation to solve

Answer 15

Particular integral function for \[e ^{-x}\]
is
\[y=Ce ^{-x}\]

but since you already have the form \[Ae ^{-x}\] in your complimentary function you must multiply our particular integral by a factor of x to get

\[y=Cxe ^{-x}\]

from here you will differentiate it twice

then substitute y, first derivative, second derivative into your original second ODE

you will then solve for C

you will then substitute C into your particular integral to get Y

for example, if C=3

you get

\[y=3e ^{-x}\]


this is added to your complimentary function

next you have to do the same method with the particular integral for cos(x)

particular integral for cos(3x) is \[y= C _{1}sinx + C_{2}cosx\]

once you work out this particular integral, add this onto your complimentary function


btw your complimentary function is

\[y=Ae^{2x} + Be^{x}\]

your general solution is \[y=Ae^{2x} + Be^{x}\] + particular integral for \[e^{-x}\] + particular integral for cos(3x)

Answer 16

I feel like that is explained pretty poorly

Answer 17

no i was going over what you said and im inputting the information to see if i can get it. thanks im going to try that out thanks.