I have the following equation
d^2y/dx^2 +4dy/dx +8y = 4sin(2x) - 12cos(2x) y(0)=1 & y'(0)=-6
For the auxillary equation I have (m + 2)^2 = sqrt -4
which gives m=-2-2i & m=-2+2i
Which gives y=e^-2x(Ccos(-2x) + Dsin(-2x))
Now to tackle particular integral.
Try y=acos(2x) +...
Thanks for your help everyone, my revised solution works! I've learned a lot, your gentle pointers eventually made the penny drop.
At least now I know how to check my solutions!
Thanks again.
James
P.S
I'll be moving onto 2nd order differential equations next and looking forward...
Thanks for your help Mark, I've been over my solution several times but always get the same - and its worng? I just can't see where I've gone wrong...
Is it my integrating factor?
James
Yes my solution satisfies condition y(0)=1
So I now need to differntiate my solution and substitute back into the equation? I'm not sure I get what you mean sorry? Bear in mind I'm trying to learn this...! lol
James
On checking my work I think I've mad a mistake
the equation should be
y(x) = 1/IF{integral of 2x/(x^2 + 1)}
Which makes it different...
My final revised answer is
y(x) = (x^2 + 1)(ln{x^2 + 1}) + C
After plugging in values y(0)=1
I have y(x)=(x^2 + 1)(ln{x^2 + 1}) +...
Integrating factor!
As promised I'm back with integrating factor differential equation.(x^2 + 1)dy/dx -2xy = 2x(x^2+1) y(0)=1
First put into standard from by dividing thru by (x^2 +1 )dy/dx -2xy/(x^2 + 1) = 2x
Integrating factor is given by exp( integral of -2x(x^2 + 1))...
Ok everyone I've had a rethink on my integration...
Here goes.
Integral of 1/(1 + 4t^2) becomes
1/4 Integral of 1/(1/4 + t^2)
a^2= 1/4 so a=1/2
Standard integral 1/a arctan(x/a)
So finally after integrating I arrive at
1/2 arctan(2t)
What do you think?
James
Can you show me how I dropped the factor of 1/2?
I have 1/(4t^2 + 1)
If x=2t then x^2=4t^2
So using 1/(x^2 + a^2) where a=1=a^2
I'm struggling with it really!
Cheers James