Question

solve the boundary value problem if possible.
\[y''-3y'+2y=0\]
y(0)=1 y(3)=0
auxiliary equation
\[r^2-3r+2=0\]
\[(r-2)(r-1)=0\]
\[r_1=2\]
\[r_2=1\]
\[y(x)=c_1e^{2x}+c_2e^{x}\]
\[y(0)=c_1e^{2(0)}+c_2e^0=1\]
\[y(0)=c_1+c_2=1\]
\[y(3)=c_1e^6+c_2e^3=0\]
\[e^6-e^6c_2+c_2e^3=0\]
\[-c_2+c_2\frac1{e^3}=0\]
\[\frac1{e^3}=1\]

Answer 1

\[1-c_2+c_2\frac1{e^3}=0\]

Answer 2

everything looked fine until this
y(0)=c1+c2=1

y(3)=c1e^6+c2e^3=0

c1+c2=1
c1*e^6+c2e^3=0

2 equations 2 unknowns

solve for c1 and c2

e^6 and e^3 are both constants

Answer 3

one moment...

Answer 4

e^6=403.428555
e^3=20.08553692

c1+c2=1
403.429*c1+20.086*c2=0

c1=1-c2

403.429(1-c2)+20.086c2 =0

403.429-403.429c2+20.086c2=0

403.329=(403.429-20.086)c2

Answer 5

\[c_2=\frac1{1-e^{\frac13}}\]

Answer 6

\[c_1=1-\frac1{1-e^{\frac13}}\]

Answer 7

\[c2=\frac{e^6}{e^6-e^3}\]
if you divided top and bottom by e^6
wont you get

\[c2=\frac{1}{1-e^{-3}}\]
or did i get my math wrong
@MathSofiya

Answer 8

oh just realized you were the one asking the question

Answer 9

If the statement below is correct
\[e^6-e^6c_2+c_2e^3=0\]
then wouldn't we get
\[1-c_2+c_2\frac1{e^3}=0\]
since
\[\frac{e^3}{e^6}=\frac1{e^3}\]

Answer 10

where does the e^1/2 come from?

Answer 11

meh ill re do it using actual exponents

Answer 12

what do you mean e^.5?

Answer 13

i mean you answer for c2

Answer 14

\[1-c_2+c_2\frac1{e^3}=0\]
\[1=c_2-c_2\frac1{e^3}\]
\[1=c_2(1-\frac1{e^3})\]

Answer 15

then solve for c_2
\[c_2=\frac1{1-e^{\frac13}}\]

Answer 16

\[c_1+c_2=1\]
\[c_1e^6+c_2e^3=0\]

\[c_1=1-c_2\]
\[(1-c_2)e^6+c_2e^3=0\]
\[e^6-c_2e^6+c_2e^3=0\]
\[e^6=c_2(e^6-e^3)\]
\[c_2=\frac{e^6}{e^6-e^3}\]
\[c_2=\frac{\frac{e^6}{e^6}}{\frac{e^6-e^3}{e^6}}\]
\[c_2=\frac{1}{1-\frac{1}{e^3}}\]

\[\frac{1}{e^3}=e^{-3}\]
\[c_2=\frac{1}{1-e^{-3}}\]

Answer 17

so I was right?

Answer 18

c_1=1-\frac1{1-e^{\frac13}}

Answer 19

\[c_1=1-\frac1{1-e^{\frac13}}\]

Answer 20

and
\[c_2=\frac1{1-e^{\frac13}}\]

Answer 21

using your last step before you solved for c2
\[1=c_2(1−\frac{1}{e^3})\]
\[1=c_2(1−e^{-3})\]
\[c_2\frac{1}{(1−e^{-3})}\]

Answer 22

e^-3 is very different from e^(1/2)

Answer 23

\[e^{\frac{1}{2}}=\sqrt{e}\]

Answer 24

that's when I made a mistake and I corrected it
My mistake was \[-c_2+c_2\frac1{e^3}=0\]
and it should have been
\[1-c_2+c_2\frac1{e^3}=0\]

Answer 25

your final solution shouldnt have e^(1/2)

assuming i didnt read the wrong problem

Answer 26

scroll up I a little I have my solutions for c_1 and c_2
about 5 posts up

Answer 27

i saw it and im saying its incorrect....
the 1/2 should be -3

Answer 28

\[c_1=1-\frac1{1-e^{\frac13}}\]
\[c_2=\frac1{1-e^{\frac13}}\]
I've posted them also 24 minutes ago

Answer 29

the only way you get 1/2 from 1/e^3 is if you had them to the -1/6th power

Answer 30

where are you getting 1/2 from?

Answer 31

sorry i mean it shouldnt be e^1/3

Answer 32

the 3 looks like a 2 too small

Answer 33

oh I see...why not though?

Answer 34

@satellite73 what do you think?

Answer 35

\[\frac{1}{e^3}=e^{-3}\]
\[e^{\frac{1}{3}}=\sqrt[3]{e}\]

Answer 36

\[1−c_2+c_2\frac{1}{e^3}=0\]
\[1=c_2-c_2\frac{1}{e^3}\]
\[1=c_2(1-\frac{1}{e^3})\]
\[c_2=\frac{1}{1-\frac{1}{e^3}}\]
\[c_2=\frac{1}{1- e^{-3}}\]

Answer 37

are we agreeing? I posted the question again to the left since this post is getting a little long

Answer 38

\[1−c_2+c_2\frac{1}{e^3}=0\]
\[1−c_2+c_2e^{-3}=0\]

Answer 39

I think we come to the same conclusion, I think the main confusion was the 1/3 looking like a 1/2

Answer 40

ill just say it
c2 is equal to 1 over (1 minus e to the negative 3rd power)

Answer 41

i keep seeing your answer as

c2 is equal to 1 over (1 minus e to the one-third power)

Answer 42

what's the difference?

Answer 43

e to the one third is not equal to e to the negative third

whenever you have fractions in exponents, it means the root of

x^(1/2)= the square root of x
x^(1/3)= the third root of x

x^-1 = 1 over x
x^-2= 1 over x^2

if x=2

x^(1/2)=2
x^-2= 1/4

2 is not equal to 1/4