# differential equations 2

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• Jul 9th 2007, 10:06 AM
emily28
differential equations 2
just thought that i would post the last 4 differential equations because these are the last 4 i dont know how to do http://www.mathhelpforum.com/math-he...cons/icon9.gif

1) (d2y/dx2)-3(dy/dx)+2y=0

2) (d2y/dx2)-3(dy/dx)+2y=e^5x

3) (d2y/dx2)-3(dy/dx)+2y=e^x

they all work on eachother so shouldnt be that hard...right?!http://www.mathhelpforum.com/math-he...cons/icon7.gif
• Jul 9th 2007, 10:11 AM
ThePerfectHacker
Quote:

Originally Posted by emily28
1) (d2y/dx2)-3(dy/dx)+2y=0

$k^2 - 3k + 2=0$
$k=-1,-2$
Thus,
$y=C_1e^x+C_2e^{2x}$

Quote:

2) (d2y/dx2)-3(dy/dx)+2y=e^5x
Look for particular of form,
$y = Ae^{5x}$

Quote:

3) (d2y/dx2)-3(dy/dx)+2y=e^x
Look for particular of form,
$y=Axe^x$

Quote:

4) (d2y/dx2)-3(dy/dx)+2y=xe^2x
Look for particular of form,
$y=Ax^2e^x$
• Jul 10th 2007, 08:16 AM
emily28
Quote:
2) (d2y/dx2)-3(dy/dx)+2y=e^5x
Look for particular of form,
http://www.mathhelpforum.com/math-he...8cf2fb9a-1.gif

i have used the particular solution of $y=Cxe^{5x}$

so
$y' = Ce^{5x} + 5Cxe^{5x}$

and $y'' = 5Ce^{5x} + 5Ce^{5x} +25 Cxe^{5x}$

which is $Ce^{5x}(25x+10)$

then $Ce^{5x}(25x+10)$ -3 $Ce^{5x} + 5Cxe^{5x}$ +2Cxe^{5x} = e^{5x}

giving 10c - 3c = 1
and c=1/7

so the general solution is $y= Ae^x + Be^{2x} + 1/7 xe^{5x}$

is this right?
• Jul 10th 2007, 08:20 AM
emily28
Quote:
3) (d2y/dx2)-3(dy/dx)+2y=e^x
Look for particular of form,
http://www.mathhelpforum.com/math-he...73f8d358-1.gif

i have used the particular solution of $y=Cxe^{x}$

so
$y' = Ce^{x} + Cxe^{x}$

and $y'' = Ce^{x} + Ce^{x} + Cxe^{x}$

which is $Ce^{x}(x+2)$

then $Ce^{x}(x+2)$ -3 $Ce^{x} + Cxe^{x}$ +2Cxe^{x} = e^{x}

giving 2c - 3c = 1
and c=-1

so the general solution is $y= Ae^x + Be^{2x} - xe^{x}$

is this right?
• Jul 11th 2007, 09:18 AM
emily28
are my previous attempts right and does anyone know where i have gone wrong on the last attempt? any advice would be appreciated:o
• Jul 11th 2007, 03:14 PM
emily28
please could someone tell me if ive got the previous attempts right? it would really help
• Jul 11th 2007, 06:10 PM
Jhevon
Quote:

Originally Posted by emily28
Quote:
2) (d2y/dx2)-3(dy/dx)+2y=e^5x
Look for particular of form,
http://www.mathhelpforum.com/math-he...8cf2fb9a-1.gif

i have used the particular solution of $y=Cxe^{5x}$

so
$y' = Ce^{5x} + 5Cxe^{5x}$

and $y'' = 5Ce^{5x} + 5Ce^{5x} +25 Cxe^{5x}$

which is $Ce^{5x}(25x+10)$

then $Ce^{5x}(25x+10)$ -3 $Ce^{5x} + 5Cxe^{5x}$ +2Cxe^{5x} = e^{5x}

giving 10c - 3c = 1
and c=1/7

so the general solution is $y= Ae^x + Be^{2x} + 1/7 xe^{5x}$

is this right?

this solution is not correct. why didn't you use $y = Ae^{5x}$ as suggested?
• Jul 11th 2007, 06:20 PM
Jhevon
Quote:

Originally Posted by emily28
Quote:
3) (d2y/dx2)-3(dy/dx)+2y=e^x
Look for particular of form,
http://www.mathhelpforum.com/math-he...73f8d358-1.gif

i have used the particular solution of $y=Cxe^{x}$

so
$y' = Ce^{x} + Cxe^{x}$

and $y'' = Ce^{x} + Ce^{x} + Cxe^{x}$

which is $Ce^{x}(x+2)$

then $Ce^{x}(x+2)$ -3 $Ce^{x} + Cxe^{x}$ +2Cxe^{x} = e^{x}

giving 2c - 3c = 1
and c=-1

so the general solution is $y= Ae^x + Be^{2x} - xe^{x}$

is this right?

this solution is correct.

good job :)
• Jul 12th 2007, 02:43 AM
emily28
cheers its good to know one of them was right, ill have another attempt at 2) and Jhevon do you know if what i have done of my final one is correct?
• Jul 12th 2007, 04:33 AM
emily28
looking at

2) (d2y/dx2)-3(dy/dx)+2y=e^5x
Look for particular of form,
http://www.mathhelpforum.com/math-he...8cf2fb9a-1.gif

$y' = 5Ce^{5x}$

and $25Ce^{5x}$

then $25Ce^{5x} - 3(5Ce^{5x}) + 2(Ce^{5x}) = e^{5x}$

giving 12c = 1
and c=1/12

so the general solution is $y= Ae^x + Be^{2x} + 1/12 e^{5x}$

is this right now ?
• Jul 12th 2007, 08:04 AM
ThePerfectHacker
Quote:

Originally Posted by emily28
looking at

2) (d2y/dx2)-3(dy/dx)+2y=e^5x
Look for particular of form,
http://www.mathhelpforum.com/math-he...8cf2fb9a-1.gif

$y' = 5Ce^{5x}$

and $25Ce^{5x}$

then $25Ce^{5x} - 3(5Ce^{5x}) + 2(Ce^{5x}) = e^{5x}$

giving 12c = 1
and c=1/12

so the general solution is $y= Ae^x + Be^{2x} + 1/12 e^{5x}$

is this right now ?

Look right to me.