# Thread: 2nd order diff eq

1. ## 2nd order diff eq

im having problems again with the constant at the end!! just need a bit of help trying to figure it out..

$\displaystyle \begin{gathered} \frac{{d^2 y}} {{dx^2 }} - 8\frac{{dy}} {{dx}} + 7y = 0 \hfill \\ \hfill \\ (\lambda - 1)(\lambda - 7) \hfill \\ \lambda = 1,\lambda = 7 \hfill \\ \hfill \\ y(x) = Ae^x + Be^7 ^x \hfill \\ \hfill \\ 0 = y(0) = A + B \hfill \\ \hfill \\ 5 = y'(0) = ... \hfill \\ \end{gathered}$

Its just this last part with the constant here I'm strugging with can someone help finish this off..

I know the answer it should be,

$\displaystyle y = \frac{5} {6}(e^7 ^x - e^x ))$

2. Originally Posted by dankelly07
im having problems again with the constant at the end!! just need a bit of help trying to figure it out..

$\displaystyle \frac{d^2 y}{dx^2 } - 8 \frac{dy}{dx} + 7y = 0$

$\displaystyle (\lambda - 1)(\lambda - 7) = 0$

$\displaystyle \lambda = 1 , \lambda = 7$

$\displaystyle y(x) = Ae^x + Be^{7x}$

$\displaystyle 0 = A + B$

$\displaystyle 5 = y'(0)$.....

Its just this last part with the constant here I'm strugging with can someone help finish this off..

I know the answer it should be,

$\displaystyle y = \frac{5} {6}(e^{7x} - e^x )$

find a better program to output your latex, whatever you just used sucked!

firsly you must find out what $\displaystyle \frac{dy}{dx}$ is

$\displaystyle \frac{dy}{dx} = Ae^x + 7Be^{7x}$

then sub in the given values

$\displaystyle 5 = A + 7B \ \ \ (1)$
$\displaystyle 0 = A + B \ \ \ \ (2)$

and the rest is easy.

Bobak

3. Thanks for your help, any suggestions on a better latex program? that was the first time i used it...

4. Well I did all the corrections by hand on that. If your going to post here frequently it may be worth picking up a few common inputs such as \frac \int

Look on the Latex Help forum of the website http://www.mathhelpforum.com/math-help/latex-help/ there are many posts about programs.

Bobak

5. ## similar question

ok now i have another problem...

say for example I have a general solution of

y(x) = Aexp{-4x}+Bex{-x}

with y(0)=1 and y(1)=0

I get

1 = A + B
0 = Aexp{-4}+Bexp{-x}

so if Iam right how do I procced from here?

6. Originally Posted by dankelly07
ok now i have another problem...

say for example I have a general solution of

y(x) = Aexp{-4x}+Bex{-x}

with y(0)=1 and y(1)=0

I get

1 = A + B
0 = A exp{-4}+B exp{-1} Mr F says: Note the correction.

so if Iam right how do I procced from here?
Solve the equations simultaneously for A and B. $\displaystyle e^{-4}$ and $\displaystyle e^{-1}$ are just numbers ......

7. thanks, I just want to check,
so now I get

-exp{-1}=-exp{-1}
Aexp{-4}=0

so

Aexp{-3} = 0

A= exp{-3}

B=-exp{-5} ???

8. Originally Posted by dankelly07
thanks, I just want to check,
so now I get

-exp{-1}=-exp{-1}
Aexp{-4}=0

so

Aexp{-3} = 0

A= exp{-3}

B=-exp{-5} ???

Sorry but you're working makes absolutely no sense.

And did you check your answers? Do they satisfy A + B = 1?

If I asked you to solve

1 = A + B

0 = 2A + 3B

how would you go about doing it .......?

9. I would minus 3 from the first equation to get rid of the B to get

-A = 1

A + B = 1
2A + 3B = 0

A = -1
B = 1

......
my maths history isnt gr8 as u can probably tell...

10. Originally Posted by dankelly07
I would minus 3 from the first equation to get rid of the B to get

-A = 1

A + B = 1
2A + 3B = 0

A = -1
B = 1

......
my maths history isnt gr8 as u can probably tell...
If this is a true representation of your level of algebraic skill, your prospects for success in a subject covering differential equations are very bleak unless you embark on an extensive revision program that covers basic algebra.

11. well I guess thats what I get for studying Maths at uni without studying maths at college... I studied music at college, which doesnt really help..lol thanks anyways