# Math Help - differential equations 2

1. ## 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

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?!

2. Originally Posted by emily28
1) (d2y/dx2)-3(dy/dx)+2y=0
$k^2 - 3k + 2=0$
$k=-1,-2$
Thus,
$y=C_1e^x+C_2e^{2x}$

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

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

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

3. Quote:
2) (d2y/dx2)-3(dy/dx)+2y=e^5x
Look for particular of form,

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?

4. Quote:
3) (d2y/dx2)-3(dy/dx)+2y=e^x
Look for particular of form,

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?

5. are my previous attempts right and does anyone know where i have gone wrong on the last attempt? any advice would be appreciated

6. please could someone tell me if ive got the previous attempts right? it would really help

7. Originally Posted by emily28
Quote:
2) (d2y/dx2)-3(dy/dx)+2y=e^5x
Look for particular of form,

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?

8. Originally Posted by emily28
Quote:
3) (d2y/dx2)-3(dy/dx)+2y=e^x
Look for particular of form,

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

9. 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?

10. looking at

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

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

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 ?

11. Originally Posted by emily28
looking at

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

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

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.