Question

Help me for a second time to answer my problem

Elimination of Constants

y=Cx+C^2+1
y=1+C1e^x+C2e^-2x
y=9x^2 +Bx+c

Answer 1

@ganeshie8 Can you please help me for a second time?

Answer 2

check the first equation once.... looks there is a typo ?

Answer 3

ganeshie8 there is no typo that is different problems :)

Answer 4

@dan815 Help me again? :)

Answer 5

go about it the same way as before
find y':
and y'':
Then try to find a linear combination where the constants dissappear

Answer 6

y=Cx+C^2+1

y' = ?

Answer 7

You're right I think my professor has a typo.

Answer 8

keep in mind, C is just a constant

Answer 9

are you sure its not cx^2 vanessa?

Answer 10

dan815 yes i'm sure

Answer 11

ok continue

Answer 12

I think the x in Cx is a sub

Answer 13

it helps if you can take a screenshot of equations and attach...

Answer 14

@ganeshie8 Here's the photo

Answer 15

that doesn't help much lol both look same !

Answer 16

lets skip this and work the remaining problems first... ?

Answer 17

What problems?

Answer 18

remaining two problems :

y=1+C1e^x+C2e^-2x
y=9x^2 +Bx+c

Answer 19

start by creating the table :

y = 1+C1e^x+C2e^-2x
y' = ?
y'' = ?

Answer 20

see if can you find the first and second derivatives ^^

Answer 21

Ahm ok :)

Answer 22

HAHAHA I can derived that XD Lol

Answer 23

good, find the derivatives :)

Answer 24

*I can't XD

Answer 25

B'coz C is a constant XD

Answer 26

lol that makes our life easy

Answer 27

you can treat C as a number while differentiating

Answer 28

hahaha You're right XD

Answer 29

give it a try

Answer 30

y = 1+C1e^x+C2e^-2x
y' = (1+C1e^x+C2e^-2x)'

Answer 31

whats the derivative of 1 ?

Answer 32

it's become 0 XD

Answer 33

see you do know how to differentiate it :) finish it off... !

Answer 34

y = 1+C1e^x+C2e^-2x
y' = (1+C1e^x+C2e^-2x)'
= 0 + ?

Answer 35

Lol I just can diffirentiate only 1 XD

Answer 36

the game is over XD

Answer 37

game just started :P

Answer 38

wahahaha So Solve it ! XD Don't pressure me XD

Answer 39

derivative of e^x is also easy :

\(\large (e^x)' = e^x\)

Answer 40

e^x is the only function whose derivative equals the function itself.

Answer 41

Then? What's next ? :D

Answer 42

lol this is your game - you will need to sweat :P

Answer 43

Also \(\large (e^{ax})' = ae^{ax} \)

Answer 44

Wahaah Lol

Answer 45

use them to find y' :


y = 1+C1e^x+C2e^-2x
y' = (1+C1e^x+C2e^-2x)'
= 0 + ?

Answer 46

Can we subtract the 1 and 2 eq. ? XD

Answer 47

(y)2 ?
y=2C1e^x+2C2e^-2x
Y'=C1e^x+(-2C2e^-2x)

Answer 48

y'-y=C1e^x

Answer 49

Am I right? @ganeshie8

Answer 50

y' is right !

Answer 51

lets not rush to the final answer yet, cuz we have two constants c1 and c2.. that tells us that we need to differentiate it twice

Answer 52

Ahh ok ! :)

Answer 53

y = 1+C1e^x+C2e^-2x
y' = 0 + C1e^x -2C2e^-2x = C1e^x - 2C2e^-2x
y'' = ?

Answer 54

differentiate y' again ^

Answer 55

y' = C1e^x - 2C2e^-2x
y'' = ?