Question

Find the general solution to the differential equation:


d^2y/dx^2 - 2(dy/dx)+4y = 2+3x+sinx

Answer 1

First you have to solve the auxiliary equation.
That will be:
\[\lambda^2-2\lambda+4=0\]

Answer 2

This one is a nasty one as it does not have real roots, but that is not a problem.

Answer 3

I have to go for 5 min now but I will continue later if you wish

Answer 4

thanks mate for an assignment worth a few marks would be appreciated.

Answer 5

So I'm back.
Solving the above equation for lamdba needs complex numbers but I guess that you know that.
Solutions are:
\[1+i \sqrt{3}\]
\[1-i \sqrt{3}\]

Answer 6

yep

Answer 7

If the auxiliary equation has complex roots alfa± i, then two linearly independent solutions
to ay'' + by' + cy = 0 are
e^alfa t cos t and e^alfa t sin t,

Answer 8

So the general solution is:
\[Ae^tcos \sqrt{3}t + Be^tsin \sqrt{3}t\]

Answer 9

This is just the 0 part

Answer 10

All the t-s should be x

Answer 11

2+3x+sinx for this you can do it in two parts
first for the 2+3x
d^2y/dx^2 - 2(dy/dx)+4y

-3/4 x^2+1/8

Answer 12

If I am right but I need to double check it, just a sec my girlfriend is calling me.

Answer 13

no I am stupid.

Answer 14

no your not thanks you are on the right path

Answer 15

it has to be in the form of ax+b
than y''=0
y'=a
y=ax+b
-2a+4(ax+b)=2+3x
4ax-2a+4b=2+3x
a=3/4
b=8/7

Answer 16

So it will have a part: 3/4x+8/7
Can you check this while I make a tea :-)

Answer 17

yeah no worries

Answer 18

OK for the sinx bit
y has to be in the form of y= Qsinx+Wcosx
than y'=Qcosx-Wsinx
y''=-Qsinx-Wcosx

Answer 19

-Qsinx-Wcosx-2(Qcosx-Wsinx)+4(Qsinx+Wcosx)=sinx
3Wcosx-2Qcosx+3Qsinx+2Wsinx=sinx
3W=2Q
3Q+2W=1
Q=3/2 W
3(3/2 W)+2W=1
13/2W=1
W=2/13
Q=3/13

Answer 20

And we are done. The solution is the general sol above + the linear bit+the sin/cos bit

Answer 21

Happy? :-)

Answer 22

A and B from the general solution are constants. If more info is given than you can calculate it

Answer 23

Andras thankyou very much.

Answer 24

just one thing is your 8/7 correct or should it be 7/8 ?

Answer 25

ur welcome, hope you get a few points! What course is it?

Answer 26

not sure, you should do the calculation too. On the computer I dont see it through so well.

Answer 27

Engineering Maths III
I will work through my notes and your answers. Thanks again

Answer 28

University In Australia

Answer 29

Andras if possible could you please explain this part, maybe show all the working, I keep getting stuck. I am not quite sure of the algebra. Thanks

it has to be in the form of ax+b
than y''=0
y'=a
y=ax+b
-2a+4(ax+b)=2+3x
4ax-2a+4b=2+3x
a=3/4
b=8/7

Answer 30

In the equation you have 2+3x
This can only come from some kind of polynomial. If it has x^2 or higher power than that part cannot disappear. So the polynomial has to be ax+b (a and b are constants)
Do you need help for solving it?

Answer 31

Andras, thanks for your reply, I was able to work through it myself. Its all good now.

Answer 32

OK, great!