Question

ok need a little help :
need to solve these--\[\large \frac{ d^2x }{ dt^2 }-y=2x+2t\\\large \frac{ d^2x }{ dt^2 }+4\frac{ dy }{ dt }=3y\]

Answer 1

i was trying to solve it using CF and PI

Answer 2

do you have notes on this

Answer 3

my work:
\[y=\frac{ d^2x }{ dt^2 }-2x-2t\]
now putting this one in the second equation --
\[\frac{ d^2x }{ dt^2 }+4[\frac{ d^3x }{ dt^3 }-2\frac{ dx }{ dt }-2]=3[\frac{ d^2x }{ dt^2 }-2x-2t]\\2\frac{ d^3x }{ dt^3 }-\frac{ d^2x }{ dt^2 }-4\frac{ dx }{ dt }+3x=4-3t\]
so the auxiliary eq. will be
\[(2D^3-D^2-4D+3)x=4-3t\\2m^3-m^2-4m+3=0\]

Answer 4

solving it
\[(m-1)^2(2m+3)=0\\m=1,1,-3/2\\so~~C.F=(c_1+c_2t)e^t+c_3e^{\frac{ -3 }{ 2 }t}\]
am i right so far?

Answer 5

hartnn?

Answer 6

looks good,
bit rusty on solving D.Es,
recently only using laplace transform to solve DEs

Answer 7

yeah LT is easier way, but i have to use these
now im going ahead with the PI

Answer 8

we can always verify the answer later using LT

Answer 9

now in the proper integral part--\[P.I=\frac{ 1 }{ 2D^3-D^2-4D +3}(4-3t)\]
right?

Answer 10

now what to do in the next step ,i dont see any way to use some formulaes :(

Answer 11

use partial fractions ?
1/(D-1)^2 (2D+3) = A/(D-1) + B/(D-1)^2 + C/(2D+3)

Answer 12

then apply each of them individually to 4-3t

Answer 13

yes good idea i can use partial or binomial ,ok wait a min

Answer 14

\[PI=\frac{ 1 }{ 3 }[1+\frac{ 2D^3-D^2-4D }{ 3}]^{-1}(4-3t)\\=\frac{ 1 }{ 3 }[1-\frac{ 2 }{ 3 }D^3+\frac{ D^2 }{ 3 }+\frac{ 4 }{ 3}D+....](4-3t)\]
ok?

Answer 15

right, yes

Answer 16

you can drop D^2 and D^3 terms...

Answer 17

yeah it gives me
\[PI=1/3(4-3t-4)=-t\]

Answer 18

yeah
\[so\\x=(c_1+c_2t)e^t+c_3e^{\frac{ -3 }{ 2 }t}-t\]
thats the answer thanks :)

Answer 19

welcome ^_^

Answer 20

lets try LT

Answer 21

\[first~eq\\s^2X(s)-Y(s)=2X(s)+2/s^2............(1)\\second~one\\s^2X(s)+4Y(s)=3Y(s)...............(2)\]
uhh i think it will more complicated

Answer 22

*mistake
\[second ~one\\s^2X(s)+4sY(s)=3Y(s)\]

Answer 23

right, more complicated than the binomial method u used,
less if you had used partial fractions as i suggested...

Answer 24

@sidsiddhartha is this a system of ODEs, or two separate ones?

Answer 25

yes system of ODEs @SithsAndGiggles

Answer 26

Ok, the reason I ask is because you're only given one second order derivative. I'm not familiar with the methods, or perhaps their names (PI and CF?). It looks like one of them might be some form of elimination where you substitute one eq. into the other.

Another approach would be cumbersome, but it should work. It would involve order-reduction of the given system. Let
\[\begin{pmatrix}x\\y\\x'\\y'\end{pmatrix}=\begin{pmatrix}u_1\\u_2\\u_3\\u_4\end{pmatrix}\]
Then you have
\[\begin{cases}
x''=2x+y+2t\\
x''+4y'=3y&\text{(are there really two of }x''?)
\end{cases}\]
changed to
\[\begin{cases}
u_1'=u_3\\
u_2'=u_4\\
u_3'=2u_1+u_2+2t\overbrace{=}^{?}3u_2-4u_4\\
u_4'=???
\end{cases}\]
As you can see, though, I'm not sure how to deal with there being two \(x''\)s...

Answer 27

yeah its like separation of variables and yes that is a problem thats why i used CF and Pi method
u can check this out

Answer 28

That's a nice set of notes. You hardly see any that give exercises WITH solutions. Thanks for the upload!

Answer 29

ur welcome :)