Question

Eliminating the Arbitrary Constants in DE:

Answer 1

\[y=c _{1}+c _{2}e ^{-4x}\]

Answer 2

so i eliminated the arbitrary constant and i just want to check if my answer is correct. which is \[y \prime \prime=-4\]

Answer 3

what is y'

Answer 4

\[y \prime=-4c _{2}e ^{-4x}\]

Answer 5

\[\large y'= 0-[0\cdot e^{-4x} -4e^{-4}\cdot c_{2}] \]\[\large y'= -4c_{2}e^{-4x}\]

Answer 6

good

Answer 7

I think you can then take \(-4c_{2}\) as f(x) and \(e^{-4x}\) as g(x) and apply the product rule again.

Answer 8

...and i forgot the "x" in my first line of my last post. Supposed to be \(\large -4e^{-4x}\)

Answer 9

this is what i did

Answer 10

is that right? :/

Answer 11

Hmm...so \[\large y'' = [0\cdot e^{-4x} +[-4e^{-4x}\cdot (-4c_{2})]]\]\[\large y''= 0+(-4)(-4)c_{2}e^{-4x}\]\[\large y'' = 0+16c_{2}e^{-4x}\]

Answer 12

@kenji07 I agree with @Jhannybean work above.

Answer 13

Maybe i'm not understanding your steps in eliminating the constants. i just used the product rule but it's not an efficient way of getting rid of the constants.

Instead of writing out what you did,can you explain to us step by step what you did to solve your problem?

Answer 14

i did that first then changed my mind. how can i eliminate Csub2?

Answer 15

That's why i'm asking you to explain what you did to get rid of \(c_{1}\) and \(c_{2}\)

Answer 16

i attached it :)

Answer 17

i just manipulated the equations :\

Answer 18

I see tht you've tken thederivative to get rid of the first constant..... what did you do in the third step? \[\large y-c_{1}=c_{2}e^{-4x}\]?

Answer 19

manipulate the original equation to get c2(e^−4x)

Answer 20

c2(e^−4x)=y-c1. this is my third equation. i sustituted this in the first derivative

Answer 21

I have a feeling that's not a valid manipulation :\ Basically you're saying that you have \[\large y= c_{1} + c_{2}e^{-4x}\] then \[\large y-c_{1}=c_{2}e^{-4x}\]\[\large \frac{dy}{dx}(y-c_{1}=c_{2}e^{-4x})\]\[\large y' = -4c_{2}e^{-4x}\] So far....

Answer 22

Oh scratch out my first comment... bleh.

Answer 23

what could i do to eliminate c2 after getting te second derivative :/

Answer 24

Because now i'm not understanding your 4rth step. \[\large \text{sub 3 in 2} :y'=4(y-c_{1})\]

Answer 25

you can't get rid of all constants whatsoever

Answer 26

^ that's what i was thinking.

Answer 27

there is a negative before 4

Answer 28

Sorry about that.

Answer 29

You're stating that the function, \(y\) is dependent on the rate of change \(y'\) which is totally false.

Answer 30

the rate of change change, \(y'\), is dependent on the function \(y\) and therefore your fourth step does not make sense.

Answer 31

ok. thanks. how about this one?

Answer 32

@kenji07 That is correct.

Answer 33

That is correct. you found all your derivatives first, and found that the orignial equation equalled the same value as your original function,therefore \(y'' =y\)

Answer 34

im going to bed gn guys

Answer 35

ok. thanks guys :)

Answer 36

Ask yourself,

"What does \(y''\) mean in relation to the function \(y \)?"

Answer 37

hey. uhm. how about this one? did i get it right?

Answer 38

@Jhannybean ?

Answer 39

No that doesn't seem right.how are you factoring out a 2+3 when those arent the LCM of both terms?

Answer 40

It seems like you're trying to combine y' and y'' again and that cannot work,when you find the derivative of something you cant constantly switch between derivatives.

Answer 41

What exactly are you trying to accomplish here?

Answer 42

oh. okay. can you teach me how to solve this one?

Answer 43

still eliminating arbitrary constant

Answer 44

\[\large y=c_{1}e^{2x}+c_{2}e^{3x}\]Product rule on both. \[\large y'= [0\cdot e^{2x}+2c_{1}e^{2x}] + [0\cdot e^{3x} + 3c_{2}e^{3x}] \] simplify and use the product rule to get rid of as many terms as you can.

Answer 45

\[y \prime= 2c _{1}e ^{2x}+3c _{2}e ^{3x}\]

Answer 46

I am not going to solve the remaining derivatives for you. Just follow the steps i've taken in the other problems ive helped you solve.

Answer 47

Also, by the logic shown on the paper, you had

\[y = c_1 e^{2x} + c_2e^{3x}\]and then you waved your hands a bit and got \[y' = 5y\]but if that is true, then \[y' = 5y = 5*c_1e^{2x} + 5*c_2e^{3x}\]but differentiating \(y\) gives you \[y' = c_1e^{2x} + c_2e^{3x}\]so to match up terms, \[5c_1 = c_1\]and\[5c_2 = c_2\]which is only true if they are 0, and that's not a very interesting diffyq...

Answer 48

\[y \prime \prime=4c _{1}e ^{2x}+9c _{2}\]

Answer 49

uh, sorry, I forgot the 2 and 3 in front after differentiating, but you still end up with the same bottom line, c_1 = c_2 = 0

Answer 50

forgot the e^3x

Answer 51

please help. i really cant get it :(

Answer 52

\[y \prime \prime - y = 0\]
\[y = c _{1}e ^{2x} + c _{2}e^{3x}\]
\[y \prime = 2c _{1}e ^{2x} + 3c _{2}e^{3x}\]
\[y \prime \prime = 4c _{1}e ^{2x} + 9c _{2}e^{3x}\]
\[y \prime \prime - y = 4c _{1}e ^{2x} + 9c _{2}e^{3x} - (c _{1}e ^{2x} + c _{2}e^{3x}) = 3c _{1}e ^{2x} + 8c _{2}e^{3x}\]

Answer 53

But, since y'' - y = 0, 3c1e^2x + 8c2e^3x = 0 also!

Answer 54

Do you have an initial value for this problem? What exactly are we trying to do?

Answer 55

Oh, I see, we can substitute 3y for 3c1e^2x + 3c2e^3x

Answer 56

3c1e^2x + 8c2e^3x = 0
3y + 5c2e^3x = 0
y = (-5/3)c2e^3x

Answer 57

\[y = -\frac{ 5 }{ 3 }c _{2}e^{3x}\]

Answer 58

Still, without an initial value I'm not sure what I am doing. I guess we could go back to 3c1e^2x + 8c2e^3x = 0 and solve for x in terms of c1 and c2.

Answer 59

3c1e^2x + 8c2e^3x = 0
3c1e^2x = -8c2e^3x
3c1/(-8c2) = e^x
ln(-3c1/8c2) = x

x = ln(-3c1/8c2)

Answer 60

Kenji, what was the original problem's instructions?

Answer 61

@cebroski bro \[y \prime \prime-y=0\] is an answer to another problem. its not related to y=c1e2x+c2e3x.

Answer 62

im eliminating the arbitrary constant in y=c1e2x+c2e3x. that the original problem's instruction.

Answer 63

y=c1e2x+c2e3x
y'=2c1e2x+3c2e3x
y''=4c1e2x+9c2e3x
y'=2y+c2e3x
c2=(y'-2y)/e3x
y''=4c1e2x+9((y'-2y)/e3x)e3x
y''=4c1e2x+9(y'-2y)
c1=(y''-9(y'-2y))/4e2x
y=((y''-9(y'-2y))/4e2x)e2x+((y'-2y)/e3x)e3x
y=((y''-9(y'-2y))/4)+y'-2y
4(3y-y')=y''-9(y'-2y)
12y-4y'=y''-9y'-18y
30y+5y'-y''=0
y''-5y'-30y=0