# differential equations 2

• Jul 9th 2007, 09:06 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, 09:11 AM
ThePerfectHacker
Quote:

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

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

Quote:

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

Quote:

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

Quote:

4) (d2y/dx2)-3(dy/dx)+2y=xe^2x
Look for particular of form,
\$\displaystyle y=Ax^2e^x\$
• Jul 10th 2007, 07:16 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 \$\displaystyle y=Cxe^{5x} \$

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

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

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

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

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

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

is this right?
• Jul 10th 2007, 07:20 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 \$\displaystyle y=Cxe^{x} \$

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

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

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

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

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

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

is this right?
• Jul 11th 2007, 08:18 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, 02:14 PM
emily28
please could someone tell me if ive got the previous attempts right? it would really help
• Jul 11th 2007, 05:10 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 \$\displaystyle y=Cxe^{5x} \$

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

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

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

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

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

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

is this right?

this solution is not correct. why didn't you use \$\displaystyle y = Ae^{5x}\$ as suggested?
• Jul 11th 2007, 05:20 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 \$\displaystyle y=Cxe^{x} \$

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

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

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

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

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

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

is this right?

this solution is correct.

good job :)
• Jul 12th 2007, 01:43 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, 03:33 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

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

and \$\displaystyle 25Ce^{5x}\$

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

giving 12c = 1
and c=1/12

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

is this right now ?
• Jul 12th 2007, 07:04 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

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

and \$\displaystyle 25Ce^{5x}\$

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

giving 12c = 1
and c=1/12

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

is this right now ?

Look right to me.