Recent content by james.farrow

That's just it! I've tried putting my solution back in n it doesn't work... I'm going to have another go at it but I can't see where I've gone wrong!

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...

Hold on! I think I may have it...? I should have multiplied at all by 1/IF making my constant thus C(x^2 + 1) Or am I way off again... James

Hmmm I'm not sure, but at a guess should it be ln(C) not just C ??

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

I don't! After doing what you said I arrive at (x^2 + 1) - 1 Which is x^2. So my solution is wrong?

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

I haven't and I'm not really sure how to do it or what you mean! Forgive my ignorance but can you show me... 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))...

Thanks for your help everyone! It's appreciated! No doubt you'll be hearing from me shortly - integrating factor is next...

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

Cheers everyone! If I use the identity I still get C = pi/4. What is the check?? James