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Got it. Thanks again.

Dick, Your final remark about the more formal approach requires me to know the answer, ie g(x) already, which was obtained by the "casual" approach. So how do you find g(x) more formally than using the casual approach ? Thanks a lot !

Thank you very much.

in case I'm not being clear, the expression I posted initially is the general solution to a DE problem that I obtained (so it might be wrong !) and the question is to show that the long term behaviour as x->infinitty is that x approaches (1/2)e^(-x)

The question doesn't explicitly ask for the limit, I just assumed that's what I had to do. This is an applied maths course, not analysis. The question is asking for the long term behaviour of the solution to a differential equation...

Because I am told the answer is \frac{1}{2}e^{-x}

Homework Statement Evaluate the limit of the following as x approaches infinity \frac{e^{x}-1}{1-2e^{x}+2e^{2x}} Homework Equations The Attempt at a Solution \frac{e^{-x}-e^{-2x}}{e^{-2x}-2e^{-x}+2} which gives 0/2=0 as x approaches infinity, but apparently this is wrong.

Indeed, yes. So easy now. Thanks !

In Maple: {\it s2s}\, := \,{\frac { \left( -{\omega_{{0}}}^{2}+{\omega}^{2} \right) \sin \left( \omega\,t \right) +\cos \left( \omega\,t \right) \omega\,\lambda}{{\omega}^{4}+ \left( {\lambda}^{2}-2\,{\omega_{{0}}}^{2} \right) {\omega}^{2}\\ \mbox{}+{\omega_{{0}}}^{4}}}\]} which simplifies to...

Another look at this in Maple: [Sorry, fiddling with latex...]

no, it's not...

Yes I'm sure it's correct. This is homework so I'm loathed to post the whole details. However, take a look here: http://nuweb.neu.edu/dheiman/U600/DHO.pdf Look at the section "Driven Harmonic Oscillator" on page 4. The solution given is exactly what I want (with lambda, w_0 and F_0/m all equal...

But the coefficients of each sinusoidal term are different...

Homework Statement I have a simple harmonic oscillator system with the driving force a sinusoidal term. The question is to find the general solution and the amplitude of the steady state solution Homework Equations I found the steady state part of the solution. It is of the form...

James' equation for y is just the (explicit) general solution which follows from the solution of the characteristic equation (repeated real root). His equation for y' should have just been the derivative but he forgot to use the product rule for the term involving the constant B. Sorry, but I...

OK, more on this. Finally I think I have it. Both methods result in -y/x. In the simple method where we cancel x first, we have the proviso that x is not equal to 0 and in the method I wrote initially, my "result" also relies on x not equal to 0 (and also cos(xy) not equal to zero). Once this...

Cheer up :) It's an easy mistake to make. The main point is that you got the main method right. If I had a pound for the number of times I've made mistakes like that I'd be very rich ! See my thread on implicit differentiation if you want a laugh !

Pretty sure... No, the derivative of A.exp(-5t) is -5A.exp(-5t) and the derivative of Bt.exp(-5t) is -5Bt.exp(-5t) + B.exp(-5t)

Sorry, I am still not sure on this... Is it really valid to cancel x here ? How did cos(xy) disappear ? And why is my attempt in my initial post wrong ?

you need to differentiate the GS and plug y=1 into it. y'=B.exp(-5t) - 5(A+Bt).exp(-5t) Since A+B=0,y'(1)=B.exp(-5)=2 so B=2.exp(5) and A=-2.exp(5)

OK, I've got it now. Seems I've been led astray by the "apparent" answer.

The "apparent" answer came from http://www.analyzemath.com/calculus/Differentiation/implicit.html See the bottom of the page If the actual answer is dy/dx = -y/x, please can you explain. I showed my working in my initial posting. Where did I go wrong ?

Of course, my method is the same as hisham's but I just find it easier to follow.

This is the method I use: Take a generic linear 1st order ODE: y' + g(x)y = h(x) Then, the integrating factor I(x) = exp(int(g(x) dx)) After multiplying through by I(x) we can write: (I(x)y)' = I(x)h(x) because the integrating factor always has the property that after multiplication the LHS...

I can easily see that dy/dx=-y/x when sin(xy)=0, so I'm not sure how that helps, but thanks anyway !

The first thing you need to do is express the DE in "standard form" so that nothing is multiplying y'...so in your last example you need to multiply through by x, so it becomes y'+y/x = x. Not sure where your last equation comes from - maybe you made some typos ? Then the integrating factor will...

That looks wrong to me. Surely a(x)=I(x)=x^{-2}, giving a solution of y(x)=x^{2}(\ln(x)+C)

What do you have for I(x) ? For writing latex, click the sigma symbol on the toolbar, but I found it easier to get started using something like this http://www.codecogs.com/components/equationeditor/equationeditor.php [Broken]

Homework Statement find \frac{\mathrm{d}y}{\mathrm{d}x} where y is defined implicitly as a function of x Homework Equations x\sin(xy)=x The Attempt at a Solution x(\cos(xy)(x\frac{\mathrm{d}y}{\mathrm{d}x}+y))+\sin(xy)=1...

Thank you. That makes perfect sense !

Homework Statement A lamina of unit density consists of the region between the two curves y=\sqrt{4-x^2} and y=1-4x^2 and the x axis. Find it's moment of inertia about the x-axis. Homework Equations This is the correct answer: 2\left \{ \int_{0}^{2}\int_{0}^{\sqrt{4-x^2}}}y^2 dy dx...