Question

Differential equation help *question below* will give medal

Answer 1

So I've been working on this problem. I'm completed the first part of it and ended up with
\[y= Ae ^{4x}+Be ^{-x}-cosx\] Is this correct?

I'm not sure how to go about the second part of it with finding the equation of the curve given that it passes through the origin...

Answer 2

I just checked with Wolfram Alpha, and saw that it's supposed to be
\[y= Ae ^{4x}+Be ^{-x}-sinx\] :/

Answer 3

That n in the question should be x -.- So when x=pi

Answer 4

you are given that it passes through origin, and another point.
Thus you are given two points
(0,0) and (pi, e^(-pi) - e^(4pi))

So you can find A and B in your general solution

Answer 5

can i see your work for the first part a)

Answer 6

Sure. Can I upload a picture of it?

Answer 7

ok

Answer 8

Here it is

Answer 9

Oh here it is rotated

Answer 10

checking :) , the CF part is obviously correct

Answer 11

:D

Answer 12

looks like you did it right

Answer 13

y = acos(x) + b(sin(x)

you found that a = 0, b = -1
y = -1 sin x

Answer 14

Yes

Answer 15

y = CF + IP = Ae^(4x) + B(e^-x) + (-sin x )

Answer 16

:)
Now do I substitute the given values into the original equation given?

Answer 17

correct, substitute (0,0) and (pi, e^(-pi) - e^(4pi))

Answer 18

Alrighty. Let me do that

Answer 19

Will I end up with two equations to solve for A and B?

Answer 20

yes

Answer 21

Substituting (0,0) I ended up with 0=A+B

Answer 22

for the second point you should get
$$
\Large e^{-\pi} -e^{4\pi} = Ae^{4\pi} + B e^{-\pi} - \sin(-\pi)

$$

Answer 23

Yes, I did get that

Answer 24

ok now sin(-pi) = 0

Answer 25

for the second point you should get
$$
\Large{
e^{-\pi} -e^{4\pi} = Ae^{4\pi} + B e^{-\pi} - \sin(-\pi)
\\ \therefore \\
e^{-\pi} -e^{4\pi} = Ae^{4\pi} + B e^{-\pi} +0
}
$$

Answer 26

so A has to equal to -1, and B has to equal to 1 . do you see?

Answer 27

by equating coefficients

Answer 28

Yes, wow..I didn't even realize haha. Now I see it

Answer 29

Thank you so much! :)

Answer 30

and you can substitute the particular solutions as well into wolfram, see if that checks out

Answer 31

yes it works
http://www.wolframalpha.com/input/?i=solve+y%27%27+-3y%27+-+4y+%3D+5sin+x+%2B3cos+x+%2C+y%280%29%3D0%2C++y%28pi%29+%3D+e%5E%28-pi%29+-+e%5E%284pi%29

Answer 32

One question, on the answer sheet, the answer is:
\[y=e ^{4x}-e ^{x}-sinx\]
But if we substitute A=-1 and B=1 the answer would be different

Answer 33

I guess the answer sheet is wrong,then?

Answer 34

yes your answer sheet is wrong

Answer 35

Thank you!

Answer 36

hmmm

Answer 37

we can be absolutely sure if we plug in the solution in the original differentiation equation

Answer 38

you said the answer sheet gave you this?
$$\Large y=e ^{4x}-e ^{\color{red}{-}x}-sinx$$

Answer 39

Yes

Answer 40

No, no - for -e^x

Answer 41

oh

Answer 42

\[\Large y=e ^{4x}-e ^{\color{red}{}x}-sinx\]

Answer 43

ok checking
\[\Large y=e ^{4x}-e ^{x}-sinx\]

Answer 44

Yes, that's what the answer sheet says

Answer 45

plugging that solution into maple (computer software) gives me:
y '' - 3y ' -4y = 6*exp(x)+5*sin(x)+3*cos(x)

Answer 46

if y = exp(4*x)-exp(x)-sin(x) then
y '' - 3y ' -4y = 6*exp(x)+5*sin(x)+3*cos(x)

Answer 47

if
y = -exp(4*x)+exp(-x)-sin(x); then
y '' - 3y ' -4y = 5 sin(x) + 3 cos(x)

Answer 48

but, this is interesting

Answer 49

Alrighty. Then the answer sheet has an error

Answer 50

if
y = exp(4*x)-exp(-x)-sin(x);
then
y '' - 3y ' -4y = 5 sin(x) + 3 cos(x)

Answer 51

hmm

Answer 52

so your answer sheet has a small typo, should be -x, in the exponent there.
its interesting we got a different solution (agrees with wolfram),

Answer 53

Haha yep!

Answer 54

the solution should be unique

Answer 55

okie

Answer 56

they both satisfy x=0 , y = 0

Answer 57

Ok I did a little more work, the (unique) solution is
$$
\Large{
y=-e ^{4x}+e ^{-x}-sinx\\

}

$$
this is because the solution y(pi) = exp(-pi) - exp(4pi) is only satisfied by this.
it is not satisfied by y = e^(4x) - e^(-x) - sin (x)

Answer 58

nor is it satisfied by
y = e^(4x) - e^(-x) - sin (x)

I am confident this is the solution as you stated the problem above.

Answer 59

unless we copied the initial conditions wrong

Answer 60

you should not get two different solutions if you have two initial conditions
for 2nd order differential equation

Answer 61

I see what you're trying to say. Thank you!